** ( Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter ) **www.free-education.in is a platform where you can get pdf notes from 6th to 12th class notes, General Knowledge post, Engineering post, Career Guidelines , English Speaking Trick , How to crack interview and lots more.

## Dual Nature of Radiation and Matter Notes

### INTRODUCTION:

Do radiations have properties resembling a wave, a particle, or both? Interference, diffraction etc. prove wave nature of radiations. However, photoelectric effect verifies the particle nature of radiations. Same question go for the matters, could matters also possess wave-particle dualism? These are the main arguments we are going to discuss in this chapter notes.

Could radiation possess the wave-particle dualism?

Could the same be said about the matter?

### ROADMAP TO THE DISCOVERY OF ELECTRON:

- William Crookes, in 1870, discovered that on applying a strong electric field between cathode and anode (kept under discharge tube at low pressures), some rays were emitted by the cathode. These rays were noticed as a bright fluorescent on the glass adjacent to the cathode
- These rays were called as cathode rays by William Crookes
- In 1879, he proposed that these cathode rays are a bunch of fast moving negatively charged particles

- Then, J. Thomson (1856-1940) validated the above hypothesis. He found that when ultraviolet radiation falls on a metal surface (or when metal surface is heated), few negatively charged particles are released by the metal surface.
- He used electric and magnetic fields (across the discharge tube), mutually perpendicular to each other and the emitted electrons

- He was the first to find the values of; speed(v), and the charge to mass ratio (e/m) of the cathode ray particles, experimentally
- He found the speed of particles to be about 0.1 to 0.2 times the speed of light (), and the charge to mass ratio (e/m)
- He also found that the e/m ratio to be constant, irrespective of the material used as an emitter. This proved the universal nature of cathode ray particles.
- In 1897, he termed this negatively charged cathode ray particles as electron, and this name was used for the first time.

- A. MIlikan, in 1913 performed experiments to calculate the charge on an oil drop (popularly known as Milikan’s oil drop experiment)
- He observed that, every time the charge came to be an integral(whole number) multiple of a small charge (), which is actually the charge on an electron
- MIlikan concluded that charge on a matter is quantized (discrete)

### ELECTRON EMISSION:

- The process of emission of electrons from the metal surface when a certain amount of energy is absorbed by the metal, is called electron emission
- There are free electrons on the metal surface, present in the outermost (valence) shell that are loosely bound to the nucleus
- These electrons are emitted when a certain minimum amount of external energy is provided to the metal surface. This least value of energy is called work function of metal (denoted by Φ
_{0}) - The unit of Φ
_{0}is electron volt(eV). 1eV is defined as the energy needed to accelerate an electron through the potential difference of 1volt(V). - Work function of a material depends mainly on the nature of metal, and electronic configuration of metal (meaning the number of valence shell electrons, lesser the number, lesser the work function)

### TYPES OF ELECTRON EMISSION:

There are 3 types of electron emission; a) thermionic emission, b) field emission, and c) photoelectric emission

- Thermionic emission:

- Thermionic emission is the process in which thermal energy is used to overcome the work function of metal in order to force out the free electrons from the metal surface. It is used in thermionic converter
- In thermionic converter, comparatively hot electrode emits electrons on receiving thermal energy. These electrons move towards the comparatively cold electrode, thus producing thermionic current (and power).

- Field Emission:

- Field emission is the process where free electrons are forced out of metal surface using a strong electric field ()
- It is used in field electron microscopes, vacuum nano electronics and other such fields of study.

- Photoelectric Emission:

- Photoelectric emission is the process where free electrons are removed from a metal surface using a light of certain frequency on that metal
- It is used in image sensor (where optical image is transformed into electric signals), applied mainly in digital cameras.

- It is also used in gold leaf electroscope,used to detect and measure static electricity.

- Metal cap, on acquiring some charge passes it to the metal stem and the gold leaf inside the evacuated chamber through the instructor. The gold leaf having the same charge as the metal stem gets repelled and move away as shown in the figure.

### PHOTOELECTRIC EFFECT- A BRIEF HISTORY:

- Hertz, in 1887 discovered the phenomena of photoelectric effect for the first time
- As he was experimenting on the generation of electromagnetic waves by spark discharge, he noticed that sparks around the detector loop were intensified when ultraviolet radiation fell on the emitter plate
- Radiation falling on the metal surface provided free electrons enough energy to neutralize the attractive force of positive ions and escape themetal surface to intensify the sparks across the metal loop.

- Hallwach and Lenard during 1886-1902, researched the process of photoelectric emission
- They experimented on negatively charged zinc plate and found that on absorbing ultraviolet light zinc plate became neutral, losing its net positive charge. On further absorption of ultraviolet radiation, neutral zinc plate became negatively charged.
- This proved that electrons are released from the metal surface when ultraviolet radiation is incident on it
- They also observed that electrons are not emitted when the frequency of light incident is less than a specific least value, called the threshold frequency. Threshold frequency depends on the nature of material used as an emitter
- Some metals like zinc, magnesium etc. emitted electrons only by absorbing ultraviolet light. But, alkali metals like sodium potassium etc. could also absorb visible radiation to emit electrons, and so they were called photosensitive materials.
- So, the process in which falling of electromagnetic radiation on the metal surface results in the emission of electrons is called photoelectric effect. And the electrons released due to photoelectric effect are called photoelectrons

### EXPERIMENTAL STUDY OF PHOTOELECTRIC EFFECT:

The experimental setup consists of:

- Evacuated tube consist of photosensitive plate (emitter) and the metal plate (collector), so that electrons could freely flow from emitter to collector without any air resistance
- Photosensitive plate (emitter) to absorb visible light and emit electrons
- Metal plate (collector) to receive electrons emitted from the emitter, thus constituting a photoelectric current flow from collector plate to the emitter plate (opposite to the flow of electrons)
- Monochromatic light of short wavelength (meaning high frequency)
- Battery to accelerate emitted electrons through a potential difference
- Voltmeter to measure the potential difference between the emitter and the collector plates due to photoelectric current flow
- Ammeter to measure the value of photoelectric current.

### EXPERIMENTAL OBSERVATIONS OF PHOTOELECTRIC EFFECT:

__Variation of photoelectric current with intensity of radiation absorbed:__

- When the values of photoelectric current were plotted against the different values for intensity of light, it was observed to be a straight line passing through the origin.
- It proved that Photoelectric current, which is number of photoelectrons flowing per unit time, is directly proportional to the intensity of incident light

__Variation of photoelectric current with the potential applied:__

Case-1-When collector plate was kept at higher potential (accelerating potential) with respect to the emitter.

- As the positive potential of collector rises, photoelectric current rises for a certain period of time because the electrons emitted experience a strong attractive force by the collector
- On further increasing the positive potential, photoelectric current reaches a maximum value, beyond which it remains fixed even when positive potential is increased. This is because number of free electrons in a metal surface is always fixed.
- This maximum value of photoelectric current beyond which it remains fixed, no matter how high the positive potential gets, is called the saturation current.

Case-2-When collector plate was kept at lower potential (retarding potential) with respect to the emitter.

- As the negative potential of collector rises, photoelectric current falls. This is because the electrons emitted will experience a strong repulsive force from the collector
- As the retarding potential is further increased, even the electrons having maximum kinetic energy will be repelled by the repulsive force of the collector, and hence, the photoelectric current becomes zero
- This corresponding minimum value of retarding (negative) potential for which the photoelectric current becomes zero is called stopping potential, or cut-off potential. It is denoted by (V
_{0}). - Mathematically, stopping potential could be expressed as

Case-3-When the variation of photoelectric current was plotted against the potential for 3 different values of intensity of incident light (keeping the frequency constant).

- It was observed that stopping potential was constant for all the values of intensity. Thus, stopping potential is independent on the intensity of incident light.
- On the other hand, saturation current got higher for higher values of intensity of absorbed light

__Variation of photoelectric current with the frequency of incident light:__

- The variation of photoelectric current was plotted against the potential for 3 different values of frequency of incident light (keeping the intensity constant)
- It was observed that saturation current was constant for all the values of frequency. Thus, saturation current is independent on the frequency of incident light.
- However, stopping potential got higher in the negative direction for higher values of frequency of absorbed light

- It was also found that, after reaching a certain minimum frequency value, stopping potential followed a linear relationship with the frequency of incident light.
- This minimum value of frequency of incident light required for photoelectric emission to take place was called as threshold frequency
- The value of threshold frequency is fixed for a specific material, and it changes from one material to another

### RESULT OF THE EXPERIMENTAL STUDY OF PHOTOELECTRIC EFFECT

- Photoelectric current is directly proportional to the intensity of incident light
- Saturation current increases if intensity of radiation increases
- Stopping potential doesn’t depend on the intensity of light

- Sopping potential, and maximum kinetic energy of photoelectrons, are linearly related to the frequency of incident light
- For every material, there is a minimum value of frequency (threshold frequency) of incident light required for photoelectric emission to take place, below which no photoelectric effect occurs

- Photoelectric emission is an instantaneous process, meaning there is no time lag between the incident light and the emission of free electrons (photoelectrons).

* Exercise*: ( Dual Nature Of Radiation And Matter )

In an experiment of photoelectric effect, the intensity () was doubled. Find the change in the following:a) Photoelectric current (i), b) Stopping potential (V_{o}), and c) maximum kinetic (K_{max}) energy of the photoelectrons.

* Solution*:

a) From the experiments on photoelectric effect, we know that : i is directly propertion to I. So, photoelectric current gets doubled.

b) We know that stopping potential doesn’t depend on the intensity. So, stopping potential remains unchanged

c) We also know that maximum kinetic energy, doesn’t change with intensity. So, maximum kinetic energy remains unchanged

* Exercise*: (Dual Nature Of Radiation And Matter)

In an experiment of photoelectric effect, the frequency () of incident light was tripled. Find the change in the following: a) Saturation current (i), and b) work function of metal.

**Solution:**

a) Saturation current is independent of frequency. So, saturation current remains unchanged.

b) Work function is always fixed for a given material.

### PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT:

- As per the wave theory, the maximum kinetic energy of the photoelectron should be affected by the change in intensity. But, the experiments on photoelectric effect showed that maximum kinetic energy doesn’t depend on the change in intensity. So, this was the first inconsistency of the wave theory with the experiments.
- Wave theory didn’t talk about the relation between stopping potential and the threshold frequency.It said only increasing the intensity could overcome the stopping potential. Here came the second inconsistency of wave theory with the experimental results.
- According to the wave theory, photoelectric effect was not aninstantaneous process, it was time taking. This was the third inconsistency with the experiments.
- The above three inconsistencies showed that wave theory of light could not explain the experimentally observed characteristics of photoelectric effect.

### EINSTEIN’S PHOTOELECTRIC EQUATION:

- Since wave theory could not explain the photoelectric effect, Einstein proposed a particle theory of light for the first time
- He said that radiations are made up of specific and discrete packets of energy called as quanta of radiation energy. Each energy quantum has a value equal to
, where__hv__*h = Planck’s constant*, and*v = frequency of incident light* - These specific packets of quanta of energy are known as photons
- When a light of frequency(
*v*) (having energy*hv*) is incident on a metal surface of work function(*Φ*), 3 cases could be possible_{o} - Case-1-When (
*hv < Φ*), i.e., energy of photon is less than the work function of metal , no photoelectric emission occurs_{o}

- Case-2- When (
*hv = Φ*), i.e., energy of photon is exactly same as the work function of metal, then electrons get enough energy to just escape the metal surface._{o} - Case-3- When (
*hv > Φ*),e., energy of photon is greater than the work function of metal. Then electron, apart from getting energy to escape the metal surface, the remaining energy is provided to the electron as kinetic energy. Mathematically, it can be expressed as:_{o}

*hv = Φ _{o} + K_{max}*

Here, *K _{max }*is the maximum kinetic energy of a photoelectron

- The above equation is known as Einstein’s photoelectric equation

From the Einstein’s photoelectric equation, following points are clear:

- Photoelectric current (
*i*) is directly proportional to the intensity (*I*) of radiation. As the intensity rises, number of photons received by metal surface in a unit area per unit time rises, so number of electrons emitted rises, and hence, photoelectric current increases

* *** I ***∝ **I*

**I**

*∝*

*I*

- Since saturation current is just a maximum value of photoelectric current, saturation current gets higher with increasing intensity of incident light
- For every metal, there exists a certain minimum frequency below which no photoelectric effect occurs. This frequency is called threshold frequency.

* HV*_{O} = Φ_{O}

_{O}= Φ

_{O}

Here, *v _{o }*

*= frequency of incident light*,

*Φ*

_{o}= work function of metal- Stopping potential (
*V*) and Maximum kinetic energy (_{o}*K*) doesn’t depend upon the intensity. Because intensity is the number of photons in unit area and unit time, and the photoelectric effect take place when one electron takes one photon_{max} - Stopping potential (
*V*) and Maximum kinetic energy (_{o}*K*) is directly proportional to the frequency (_{max}*ν*)

* K*_{MAX}*∝ V V*_{O}*∝ V*

_{MAX}

_{O}

The relation between stopping potential, maximum kinetic energy and the frequency of incident light could be expressed mathematically as follows: using Einstein’s photoelectric equation

**hv = Φ _{o} + K_{max}**

**K _{max} = hν – Φ_{o}**

**Also, K _{max} = eV_{o}**

**∴ eV _{o} = hν – Φ_{o}**

On rearranging the above equation:

**V _{o} = (h/e)v + (Φ_{o}/e)**

Plotting the above equation graphically, we get:

- Photoelectric emission is an instantaneous process, meaning that there is no time gap between incident radiation and electron emission.

** Numerical Problems:** ( Dual Nature Of Radiation And Matter )

1)**Question**: Caesium metal has work function of 2.14eV. Photoelectric emission takes place when a light of frequency 6×10^{14 }Hz is incident on the metal surface. Calculate the following: a) maximum kinetic energy of the electrons emitted, b) stopping potential, and c) maximum speed of the emitted electrons.

* Solution*:

Given, Φ_{o} = 2.14eV = 2.14×10^{-19}J, and ν = 6×10^{14}Hz

a) Using Einstein’s photoelectric equation:

*hv = Φ _{o}+ K_{max}*

*K _{max} = hv – Φ_{o} = (6.6×10^{-34}×6×10^{14})J – (2.14×10^{-19})J*

*∴**K _{max} = 1.82×10^{-19}J (ans)*

b) Stopping potential is given by the equation:

*eV _{o} = K_{max}*

*V _{o} = K_{max}/e = 1.82×10^{-19} = 1.1375V (ans)*

c) Maximum speed of emitted electrons can be found using maximum kinetic energy equation:

*K _{max} = (1/2)mv_{max}^{2}*

2)**Question**: Light of wavelength 488nm is incident on an emitter plate. The photoelectrons have a stopping (cut-off) potential of 0.38V. Calculate the work function of the emitter plate.

* Solution*:

Given, *λ = 488×10 ^{-9}*m,

*V*

_{o}= 0.38VUsing Einstein’s photoelectric equation:

*hv = hc/λ = Φ _{o} + K_{max}*

and,

*K _{max} = eV_{o}*

### PHOTON PICTURE OF RADIATION:

- Radiation can behave as a wave as well as a particle under different situations. And while interacting with matter, it behaves as a particle
- Energy (
*E*) of a photon is given by:*E = hv = hc/λ*

Here, *v = frequency*, *λ = wavelength*, *c = speed of light*, and *h = Planck’s constant*

- Momentum (
*p*) of a photon is given by:*p = hv/c = h/λ*

Here symbols have their usual meaning.

- Photons of similar frequencies possess equal energies and equal momentums
- Photons are electrically neutral (not having a net positive or negative charge)
- In a photon-particle collision:

- Total energy remains constant
- Total momentum remains constant
- Total number of photons may or may not be constant.

The photoelectric effect proves the particle nature of radiation. Whereas, the phenomena like interference, diffraction, and polarization proves the wave nature of light. Hence, we can say that radiation displays both, wave nature, and particle nature, meaning a radiation possesses wave-particle dualism.

### WAVE NATURE OF MATTER: DE BROGLIE’S HYPOTHESIS:

- De Broglie proposed that if the radiations could possess dual nature, matters could also possess dual nature.
- A particle of mass (m), moving with velocity (v) could behave like a wave under suitable conditions. And the corresponding wave related to that matter is called matter wave
- De Broglie’s wavelength for matter wave is given by:
*λ = h/p*

__Case-1- (Macroscopic object)__

If we take an example of a car of mass= 900kg, moving with the velocity of 36km/hr (10m/s).

The wavelength associated with the car will be:

We can observe that the wavelength associated with the car is insignificant and can’t be detected experimentally.

Hence, for the macroscopic objects, the mass is so large that the matter wave associated with them becomes insignificant and negligible

__Case-2- (Microscopic object)__

If we take an electron, mass=9.1 X 10^{-31 }kg, moving with the speed of light (3 X 10^{8} m/s).

The wavelength associated with an electron will be:

*Kinetic energy K = (1/2)mv ^{2} = (mv)^{2}/(2m) = p^{2}/2m*

Using De-Broglie’s Hypothesis:

Putting the value of V = 50V

The wavelength associated with electron is quite large and is experimentally observable. This is because the mass of a microscopic object is very small, so wavelength becomes sufficiently large andhence, observable.

Numerical Problems:

**Question**: Monochromatic light of wavelength 632.8nm is generated by a helium-neon laser having power of 9.42mW. Evaluate the following: a) energy and momentum of each photon, b) The number of photons emitted per second, and c) speed of a hydrogen atom to have momentum equal to that of an emitted photon by the laser.

**Solution**:

Given, *λ = 632×10 ^{-9}*

*nm*,

*P = 9.42×10*

^{-3}W- Energy of a photon is given by:
**E = hv = hc/λ**

Momentum of the photon is given by: **p = hv/c = h/λ**

b. Number of photons emitted per second (n) will be given by the equation:

*Power = n×energy of a photon*

*9.42×10 ^{-3}W*

*= n×3.13×10*

^{-19 }JTo find the speed of a hydrogen atom (*v*) to have momentum same as a photon:

*v = p/m*

Here, *p = momentum of photon =1.043×10- ^{27}kgm/s*, and

*m=mass of a hydrogen atom= 1.67×10 ^{-24}kg*

**Question**: An electron has a kinetic energy of 120eV. Calculate: a) momentum, b) speed, and c) De Broglie wavelength of the electron

Solution:

Given, *K _{max} = 120×10^{-19}J*

### HEISENBERG’S UNCERTAINTY PRINCIPLE:

- This principle was in favor of the wave nature of matter
- It stated that it is impossible to simultaneously evaluate the precise position and momentum of particle. There is always some probability in predicting the position and momentum of a particle. Mathematically, it can be written as:

* (Δx)(Δp) ≥ h/(2π)*

Considering the above equation, 2 cases are possible:

- Case-1- If precise momentum(p) of an electron is known, then its wavelength by De Broglie’s hypothesis will be constant:

* λ = h/p*

It means that the wavelength has a fixed value and the wave is extended infinitely throughout the wave. Hence, it is impossible to find the position of the wave.

Mathematically, if ** p = fixed , **Then,

*Δp→0, Δx→∞*- Case-2- If the wave is localized, having finite end points

A localized wave is shown below:

As we can see in the diagram, the wavelength (λ) is not fixed, so the momentum (*p*) is also not fixed.

Hence, there is uncertainty in both, momentum (*p*) and position (*x*).

### DAVISSON AND GERMER EXPERIMENT:

- Davisson and Germer experiment showed the wave nature of electron for the first time
- They studied the diffraction effects of electron in crystal diffraction experiment
- The experimental setup for the experiment is shown below:

The experimental setup consists of:

- Evacuated chamberfor the free movement of electron without any air resistance
- Electron gunfor the emission of electron
- BatteryUsed for the acceleration of electrons inside the cylinder
- Cylinder with fine hole along its axisconnected to the battery so that electrons entering it could be accelerated to high speed.

- Nickel Targetused to deflect electron beam towards the detector
- Movable detector(collector)to detect the intensity and scattering of electrons deflected by the nickel crystal at varying voltage supplies(44 to 68 V)
- Galvanometerto measure the small values of current

Observations:

- Strong peak was detected at 55V, and the angle of scattering was observed to be 50°
- The pattern of deflected electrons was quite similar to the diffraction pattern of waves
- The wavelength corresponding to the electron (matter wave) was found to be λ = 0.165nm
- The experiment was in strong agreement with De Broglie’s hypothesis

According to De Broglie’s hypothesis, matter wave is given by:

- The experiment proved that electrons behave as a wave under specific conditions as the scattering of electron gave rise to the diffraction pattern

On putting the value of V= 55V (the value at which a strong peak was detected), we get

The wavelength calculated above using De Broglie’s hypothesis is equal to the wavelength observed in the Davisson and Germer experiment. So, Davisson and Germer experiment was consistent with De Broglie’s hypothesis.

- Hence, this experiment proved that electron behaves as a wave under specific conditions.

**CBSE Class 12 Physics Important Questions Chapter 11 – Dual Nature of Radiation and Matter**

**1 Mark Questions** ( Dual Nature Of Radiation And Matter )

**1.Calculate the energy associated in eV with a photon of wavelength ?**

**Ans. **

E = 3.09 eV

**2.Mention one physical process for the release of electron from the surface of a metal?**

**Ans.** Photoelectric emission.

**3.The maximum kinetic energy of photoelectron is 2.8 eV. What is the value of stopping potential?**

**Ans.**

**4. Calculate the threshold frequency of photon for photoelectric emission from a metal of work function 0.1eV?**

**Ans. **

**5.Ultraviolet light is incident on two photosensitive materials having work function and ( > ). In which of the case will K.E. of emitted electrons be greater? Why?**

**Ans. **

If > thus K.E. will be more for second surface whose work function is less.

**6.Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of incident radiations?**

**Ans. ** – threshold of frequency

or cut off frequency

**7. How does the stopping potential applied to a photocell change if the distance between the light source and the cathode of the cell is doubled?**

**Ans.** Stopping potential does not depend on the intensity of the light source which changes due to the change in distance from the light source.

**8.On what factor does the retarding potential of a photocell depend?**

**Ans.**It depends upon the frequency of the incident light

**9.Electron and proton are moving with same speed, which will have more wavelength?**

**Ans.** Since so electron being lighter will have more wavelengths

**2 Marks Questions** ( Dual Nature Of Radiation And Matter )

**1. Derive an expression for debroglie wavelength of an electron?**

**Ans.** If a beam of electrons traveling through a potential difference of V volt, the electron acquires kinetic energy.

Multiply by m

**2. Light of wavelength falls on an aluminum surface. In aluminum 4.2 eV are required to remove an electron. What is the kinetic energy of (a) fastest (b) the slowest photoelectron?**

**Ans. **

(a)

This is the K.E of the fastest electron

(b) Zero

**3. An electromagnetic wave of wavelength is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have de-Broglie wavelength. Prove that **

**Ans. **

Squaring we get

Or

**4. It is difficult to remove a free electron from copper than from sodium? Why?**

**Ans.** since

Where is the threshold wavelength

Since

Work function for copper is greater and it becomes difficult to remove a free electron from copper.

**5. Obtain the expression for the maximum kinetic energy of the electrons emitted from a metal surface in terms of the frequency of the incident radiation and the threshold frequency?**

**Ans. **

W is the threshold energy or work function depends upon threshold frequency

Or

**6. For a given K.E. which of the following has the smallest de–broglie wavelength: electron, proton, ?**

**Ans.** Debroglie wavelength

When E is energy

Comparing masses we get mass of is more; hence wavelength of alpha particle is minimum.

**7. Photoelectrons are emitted with a maximum speed of 710 ^{5}m/s from a surface when light of frequency 810^{14}Hz is incident on it. Find the threshold frequency for this surface?**

**Ans.**

**8. Is photoelectric emission possible at all frequencies? Give reason for your answer?**

**Ans.**No, photoelectric emission is not possible at all frequencies because it is possible only if radiation energy is greater than work function of the emitter.

**9. Assume that the frequency of the radiation incident on a metal plate is greater than its threshold frequency. How will the following change, if the incident radiation is doubled? (1) Kinetic energy of electrons**

**(2) Photoelectric current**

**Ans.**(1) If the frequency of the incident radiation is doubled is increased, hence kinetic energy is increased.

(2) If the frequency of the incident radiation is doubled there will be no change in the number of photoectrons i.e. photoelectronic current.

**10. Why are de – broglie waves associated with a moving football is not visible?**

**Ans.**The wavelength of a wave associated with a moving football is extremely small, which cannot be detected.

Since

**11. By how much would the stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from 410 ^{15} Hz to 8 10^{15 }Hz? ( )**

**Ans.** Stopping potential

**12. Work function of Na is 2.3eV. Does sodium show photoelectric emission for light on the velocity of photoelectrons?**

**Ans.**Since

Velocity of photoelectrons increases with the decrease in the wavelength of the incident light.

**13. An electron and an alpha particle have the same debroglie wavelength associated with them? How are their kinetic energies related to each other?**

**Ans.**

Dividing equation (1) and (2)

**14. An and a proton are accelerated from rest through same potential difference V. find the ratio of de–broglie wavelength associated with them?**

**Ans.**

Now kinetic energy

**15.****The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?**

**Ans.**Photoelectric cut-off voltage, = 1.5 V

The maximum kinetic energy of the emitted photoelectrons is given as:

Where,

*e*= Charge on an electronC

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is J.

**16.****The threshold frequency for a certain metal is Hz. If light of frequency Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.**

**Ans.**Threshold frequency of the metal,

Frequency of light incident on the metal,

Charge on an electron, *e*C

Planck’s constant, *h*Js

Cut-off voltage for the photoelectric emission from the metal =

The equation for the cut-off energy is given as:

Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V

**17.The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?**

**Ans.**No

Work function of the metal,

Charge on an electron, *e* C

Planck’s constant, *h*Js

Wavelength of the incident radiation, = 330 nmm

Speed of light, *c*m/s

The energy of the incident photon is given as:

It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.

** 3 Marks Questions** ( Dual Nature Of Radiation And Matter )

**1.The following table gives the values of work functions for a few sensitive metals.**

S. No. | Metal | Work function(eV) |

1. | Na | 1.92 |

2. | K | 2.15 |

3. | Mo | 4.17 |

**If each of these metals is exposed to radiations of wavelength 3300nm, which of these will not exit photoelectrons and why?**

Ans. That material will not emit photoelectrons whose work function is greater than the energy of the incident radiation.

E = 3.76 eV

Hence work function of is (4.17eV) which is greater than the energy of the incident radiation (= 3.76 eV) so will not emit photoelectrons.

**2.Define threshold wavelength for photoelectric effect? Debroglie wavelength associated with an electron associated through a potential difference V is? What will be the new wavelength when the accelerating potential is increase to 4V?**

**Ans.** The maximum wavelength of radiation needed to cause photoelectric emission is known as threshold wavelength.

Or

**3. An electron has kinetic energy equal to 100eV. Calculate (1) momentum (2) speed (3) Debroglie wavelength of the electron.**

**Ans.**

(1) (Momentum)

(2) Speed

(3) Debroglie wavelength

**4. (a) Define photoelectric work function? What is its unit?**

**(b) In a plot of photoelectric current versus anode potential, how does**

**(i) Saturation current varies with anode potential for incident radiations of different frequencies but same intensity?**

**(ii) The stopping potential varies for incident radiations of different intensities but same frequency.**

**(iii) Photoelectric current vary for different intensities but same frequency of radiations? Justify your answer in each case?**

**Ans.** (a) The minimum amount of energy required to take out an electron from the surface of metal. It is measured in electron volt (eV).

(b) (i) Saturation current depends only on the intensity of incident radiation but is independent of the frequency of incident radiation.

(ii) Stopping potential does not depend on the intensity of incident radiations.

(iii) Photoelectric current is directly proportional to the intensity of incident radiations, provided the given frequency is greater than the threshold frequency.

**5. Photoelectric work function of a metal is 1eV. Light of wavelength falls on it. What is the velocity of the effected photoelectron?**

**Ans.**

**6.The wavelength of a photon and debroglie wavelength of an electron have the same value. Show that the energy of the photon is times the kinetic energy of electron where m, c, and h have their usual meanings?**

**Ans.** Energy of a photon

Kinetic energy of an electron

But de-broglie wavelength of an electron is given by

Dividing (1) by (2)

**7.Draw a graph showing the variation of stopping potential with frequency of the incident radiations. What does the slope of the line with the frequency axis indicate. Hence define threshold frequency?**

**Ans.** Slope of the graph

Einstein photoelectric equation

Differentiating equation (1)

Thus slope is equal to the ratio of planck’s constant to the charge on electron.

Threshold frequency – The minimum values of frequency of the incident light below which photoelectric emission is not possible is called as threshold frequency.

**8.****Find the**

**(a) maximum frequency, and**

**(b) minimum wavelength of X-rays produced by 30 kV electrons.**

**Ans.**Potential of the electrons, *V*= 30 kV V

Hence, energy of the electrons, *E* eV

Where,

*e*= Charge on an electron =

**(a)**Maximum frequency produced by the X-rays =

The energy of the electrons is given by the relation:

*E* =

Where,

*h*= Planck’s constant Js

Hence, the maximum frequency of X-rays produced isHz.

**(b)**The minimum wavelength produced by the X-rays is given as:

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

**9. The work function of caesium metal is 2.14 eV. When light of frequency Hz is incident on the metal surface, photoemission of electrons occurs. What is the**

**(a) maximum kinetic energy of the emitted electrons,**

**(b) Stopping potential, and**

**(c) maximum speed of the emitted photoelectrons?**

**Ans.**Work function of caesium metal,

Frequency of light,

**(a)**The maximum kinetic energy is given by the photoelectric effect as:

Where,

*h*= Planck’s constant =

Hence, the maximum kinetic energy of the emitted electrons is

0.345 eV.

**(b)**For stopping potential, we can write the equation for kinetic energy as:

Hence, the stopping potential of the material is 0.345 V.

**(c)**Maximum speed of the emitted photoelectrons = *v*

Hence, the relation for kinetic energy can be written as:

Where,

*m*= Mass of an electron kg

Hence, the maximum speed of the emitted photoelectrons is

332.3 km/s.

**10.****The energy flux of sunlight reaching the surface of the earth is . How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.**

**Ans.**Energy flux of sunlight reaching the surface of earth,

Hence, power of sunlight per square metre, *P* W

Speed of light, *c*m/s

Planck’s constant, *h*Js

Average wavelength of photons present in sunlight,

Number of photons per square metre incident on earth per second = *n*

Hence, the equation for power can be written as:

Therefore, every second, photons are incident per square metre on earth.

**11. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be V s. Calculate the value of Planck’s constant.**

**Ans.**The slope of the cut-off voltage (*V*) versus frequency of an incident light is given as:

V is related to frequency by the equation:

Where,

*e*= Charge on an electron C

*h*= Planck’s constant

Therefore, the value of Planck’s constant is

**12.****A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?**

**Ans.**Power of the sodium lamp, *P*= 100 W

Wavelength of the emitted sodium light, = 589 nmm

Planck’s constant, *h* Js

Speed of light, *c* m/s

**(a)**The energy per photon associated with the sodium light is given as:

**(b)**Number of photons delivered to the sphere = *n*

The equation for power can be written as:

Therefore, every second, photons are delivered to the sphere.

**5 Marks Questions** ( Dual Nature Of Radiation And Matter )

**1.Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.**

**(a) Find the energy and momentum of each photon in the light beam,**

**(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and**

**(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?**

**Ans.**Wavelength of the monochromatic light, = 632.8 nm m

Power emitted by the laser, *P*= 9.42 mWW

Planck’s constant, *h*Js

Speed of light, *c*m/s

Mass of a hydrogen atom, *m*kg

**(a)**The energy of each photon is given as:

The momentum of each photon is given as:

**(b)**Number of photons arriving per second, at a target irradiated by the beam = *n*

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as:

**(c)**Momentum of the hydrogen atom is the same as the momentum of the photon,

Momentum is given as:

Where,

*v*= Speed of the hydrogen atom

**2.Light of frequency Hz is incident on a metal surface. Electrons with a maximum speed of m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?**

**Ans.**Frequency of the incident photon,

Maximum speed of the electrons, *v*m/s

Planck’s constant, *h*Js

Mass of an electron, *m*kg

For threshold frequency, the relation for kinetic energy is written as:

Therefore, the threshold frequency for the photoemission of electrons is s

**3. Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.**

**Ans.**Wavelength of light produced by the argon laser, = 488 nm

m

Stopping potential of the photoelectrons, = 0.38 V

1eV J

∴ =

Planck’s constant, *h *Js

Charge on an electron, *e*= 1.6 × 10 – 19C

Speed of light, *c*= 3 × 10 m/s

From Einstein’s photoelectric effect, we have the relation involving the work function of the material of the emitter as:

Therefore, the material with which the emitter is made has the work function of 2.16 eV.

**4.Calculate the**

**(a) momentum, and**

**(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.**

**Ans.**Potential difference, *V*= 56 V

Planck’s constant, *h*Js

Mass of an electron, *m*kg

Charge on an electron, *e*C

**(a)** At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (*v*) of each electron as:

The momentum of each accelerated electron is given as:

*p*= *mv*

kg .Therefore, the momentum of each electron is kg.

**(b)** De Broglie wavelength of an electron accelerating through a potential *V*, is given by the relation:

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.

**5.What is the**

**(a) momentum,**

**(b) speed, and**

**(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.**

**Ans.**Kinetic energy of the electron, *Ek*= 120 eV

Planck’s constant, *h*Js

Mass of an electron, *m*kg

Charge on an electron, *e*C

**(a)** For the electron, we can write the relation for kinetic energy as:

Where,

*v*= Speed of the electron

Momentum of the electron, *p*= *mv*

kg

Therefore, the momentum of the electron is kg m.

**(b)** Speed of the electron, *v*m/s

**(c)** De Broglie wavelength of an electron having a momentum *p*, is given as:

Therefore, the de Broglie wavelength of the electron is 0.112 nm.

**6.The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which**

**(a) an electron, and**

**(b) a neutron, would have the same de Broglie wavelength.**

**Ans.**Wavelength of light of a sodium line, = 589 nm m

Mass of an electron, *me*kg

Mass of a neutron, *mn*=kg

Planck’s constant, *h*Js

**(a)** For the kinetic energy *K*, of an electron accelerating with a velocity *v*, we have the relation:

We have the relation for de Broglie wavelength as:

Substituting equation (2) in equation (1), we get the relation:

Hence, the kinetic energy of the electron is J or 4.31.

**(b)** Using equation (3), we can write the relation for the kinetic energy of the neutron as:

Hence, the kinetic energy of the neutron is J or 2.36 neV.

**7.What is the de Broglie wavelength of**

**(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,**

**(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and**

**(c) a dust particle of mass kg drifting with a speed of 2.2 m/s?**

**Ans.(a)**Mass of the bullet, *m*= 0.040 kg

Speed of the bullet, *v*= 1.0 km/s = 1000 m/s

Planck’s constant, *h*= Js

De Broglie wavelength of the bullet is given by the relation:

**(b)** Mass of the ball, *m*= 0.060 kg

Speed of the ball, *v*= 1.0 m/s

De Broglie wavelength of the ball is given by the relation:

**(c)**Mass of the dust particle, *m*kg

Speed of the dust particle, *v*= 2.2 m/s

De Broglie wavelength of the dust particle is given by the relation:

s

**8. An electron and a photon each have a wavelength of 1.00 nm. Find**

**(a) their momenta,**

**(b) the energy of the photon, and**

**(c) the kinetic energy of electron.**

**Ans.**Wavelength of an electron and a photon

m

Planck’s constant, *h*=Js

**(a)** The momentum of an elementary particle is given by de Broglie relation:

It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.

**(b)** The energy of a photon is given by the relation:

Where,

Speed of light, *c*=m/s

Therefore, the energy of the photon is 1.243 keV.

**(c)** The kinetic energy (*K*) of an electron having momentum *p*,is given by the relation:

Where,

*m*= Mass of the electron = kg

*p* =

Hence, the kinetic energy of the electron is 1.51 eV.

**9.(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be****?**

**(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.**

**Ans.(a)** De Broglie wavelength of the neutron,

Mass of a neutron, *mn*= kg

Planck’s constant, *h*= Js

Kinetic energy (*K*) and velocity (*v*) are related as:

… (1)

De Broglie wavelength and velocity (*v*) are related as:

Using equation (2) in equation (1), we get:

Hence, the kinetic energy of the neutron is J or eV.

**(b)** Temperature of the neutron, *T*= 300 K

Boltzmann constant, *k*=kg

Average kinetic energy of the neutron:

The relation for the de Broglie wavelength is given as:

Where

Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

**10.Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).**

**Ans.**The momentum of a photon having energy is given as:

Where,

= Wavelength of the electromagnetic radiation

*c*= Speed of light

*h*= Planck’s constant

De Broglie wavelength of the photon is given as:

But p=mv

Where,

*m*= Mass of the photon

*v*= Velocity of the photon

Hence, it can be inferred from equations (*i*) and (*ii*) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.

**11. What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)**

**Ans.**Temperature of the nitrogen molecule, *T*= 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of the nitrogen molecule, = 28.0152 u

But 1 u = kg

∴*m*=kg

Planck’s constant, *h*=Js

Boltzmann constant, *k*=

We have the expression that relates mean kinetic energy of the nitrogen molecule with the root mean square speed as:

Hence, the de Broglie wavelength of the nitrogen molecule is given as:

Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.

**12. (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be C.**

**(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?**

**Ans.(a)**Potential difference across the evacuated tube, *V*= 500 V

Specific charge of an electron, *e/m*=

The speed of each emitted electron is given by the relation for kinetic energy as:

Therefore, the speed of each emitted electron is

**(b)**Potential of the anode, *V*= 10 MV = V

The speed of each electron is given as:

This result is wrong because nothing can move faster than light. In the above formula, the expression for energy can only be used in the non-relativistic limit, i.e., for *v* << *c*.

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:

*E*=

Where,

*m* = Relativistic mass

*m*0= Mass of the particle at rest

Kinetic energy is given as:

*K*=

**13.(a) A mono-energetic electron beam with electron speed of is subject to a magnetic field of T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals.**

**(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?**

**[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]**

**Ans.(a)**Speed of an electron, *v*= m/s

Magnetic field experienced by the electron, *B*=

Specific charge of an electron, *e/m*=

Where,

*e*= Charge on the electron =

*m*= Mass of the electron =

The force exerted on the electron is given as:

θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

The beam traces a circular path of radius, *r*. It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam.

Hence, equation (1) reduces to:

Therefore, the radius of the circular path is 22.7 cm.

**(b)** Energy of the electron beam,

The energy of the electron is given as:

This result is incorrect because nothing can move faster than light. In the above formula, the expression for energy can only be used in the non-relativistic limit, i.e., for *v* << *c*

When very high speeds are concerned, the relativistic domain comes into consideration.

In the relativistic domain, mass is given as:

s

Where,

= Mass of the particle at rest

Hence, the radius of the circular path is given as:

**14.An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure of Hg). A magnetic field of T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method. Determine e/m from the data.**

**Ans.**Potential of an anode, *V* = 100 V

Magnetic field experienced by the electrons, *B*=

Radius of the circular orbit *r*= 12.0 cm =

Mass of each electron = *m*

Charge on each electron = *e*

Velocity of each electron = *v*

The energy of each electron is equal to its kinetic energy, i.e.,

It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam. Hence, we can write:

Centripetal force = Magnetic force

Putting the value of *v *in equation (1), we get:

Therefore, the specific charge ratio (*e/m*) is

**15.(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 .What is the maximum energy of a photon in the radiation?**

**(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?**

**Ans.(a)** Wavelength produced by an X-ray tube,

Planck’s constant, *h*

Speed of light, *c*

The maximum energy of a photon is given as:

Therefore, the maximum energy of an X-ray photon is 27.6 keV.

**(b)** Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.

**16.In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two rays of equal energy. What is the wavelength associated with each -ray? ()**

**Ans.**Total energy of two -rays:

*E*= 10. 2 BeV

=

=

Hence, the energy of each -ray:

Planck’s constant,

Speed of light,

Energy is related to wavelength as:

Therefore, the wavelength associated with each -ray is

**17.Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.**

**(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.**

**(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive. Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about .**

**Ans.(a)** Power of the medium wave transmitter, *P*= 10 kW = 104 W = 104J/s

Hence, energy emitted by the transmitter per second, *E*= 104

Wavelength of the radio wave, = 500 m

The energy of the wave is given as:

Where,

*h*= Planck’s constant =

*c*= Speed of light =

Let *n* be the number of photons emitted by the transmitter.

∴*nE*1= *E*

The energy (*E*1) of a radio photon is very less, but the number of photons (*n*) emitted per second in a radio wave is very large.

The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.

**(b)** Intensity of light perceived by the human eye, *I*=

Area of a pupil, *A* = 0.4 × 10 – 4m2

Frequency of white light,

The energy emitted by a photon is given as:

*E*=

Where,

*h*= Planck’s constant =

Let *n* be the total number of photons falling per second, per unit area of the pupil.

The total energy per unit for *n *falling photons is given as:

*E*

The energy per unit area per second is the intensity of light.

s

=

The total number of photons entering the pupil per second is given as:

This number is not as large as the one found in problem **(a)**, but it is large enough for the human eye to never see the individual photons.

**18.Ultraviolet light of wavelength 2271 ¦ from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity red light of wavelength 6328 ¦ produced by a He-Ne laser?**

**Ans.**Wavelength of ultraviolet light, = 2271 ¦ =

Stopping potential of the metal, *V*0= 1.3 V

Planck’s constant, *h*

Charge on an electron, *e*=

Work function of the metal =

Frequency of light =

We have the photo-energy relation from the photoelectric effect as:

=

Let be the threshold frequency of the metal.

∴=

Wavelength of red light, =

∴Frequency of red light,

Since, the photocell will not respond to the red light produced by the laser.

**19.Monochromatic radiation of wavelength 640.2 nm from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.**

**Ans.**Wavelength of the monochromatic radiation, = 640.2 nm

=

Stopping potential of the neon lamp, = 0.54 V

Charge on an electron, *e*=

Planck’s constant, *h*

Let be the work function and be the frequency of emitted light.

We have the photo-energy relation from the photoelectric effect as:

Wavelength of the radiation emitted from an iron source, * *= 427.2 nm

*=*

Let be the new stopping potential. Hence, photo-energy is given as:

Hence, the new stopping potential is 1.50 eV.

**20.A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:**

** ,**

**The stopping voltages, respectively, were measured to be:**

**Determine the value of Planck’s constant h, the threshold frequency and work function for the material.**

**[ Note: You will notice that to get h from the data, you will need to know e (which you can take to be ). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]**

**Ans.**Einstein’s photoelectric equation is given as:

Where, = Stopping potential

*h*= Planck’s constant

*e*= Charge on an electron

= Frequency of radiation

= Work function of a material

It can be concluded from equation (1) that potential is directly proportional to frequency .

Frequency is also given by the relation:

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

The given quantities can be listed in tabular form as:

Frequency × 1014 Hz | 8.219 | 7.412 | 6.884 | 5.493 | 4.343 |

Stopping potential V0 | 1.28 | 0.95 | 0.74 | 0.16 | 0 |

The following figure shows a graph between .

It can be observed that the obtained curve is a straight line. It intersects the at Hz, which is the threshold frequency of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the * *line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line =

From equation (1), the slope can be written as:

The work function of the metal is given as:

**21.The work function for the following metals is given:**

**Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 ¦ from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?**

**Ans.**Mo and Ni will not show photoelectric emission in both cases

Wavelength for a radiation, = 3300

Speed of light, *c*=

Planck’s constant, *h*=

The energy of incident radiation is given as:

It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

**22.Light of intensity falls on a sodium photo-cell of surface area****. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?**

**Ans.**Intensity of incident light, *I*=

Surface area of a sodium photocell, *A*= =

Incident power of the light,

= =

Work function of the metal, = 2 eV

Number of layers of sodium that absorbs the incident energy, *n*= 5

We know that the effective atomic area of a sodium atom .

Hence, the number of conduction electrons in *n*layers is given as:

The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:

Time required for photoelectric emission:

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.

**23.Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 ¦, which is of the order of inter-atomic spacing in the lattice) .**

**Ans.**An X-ray probe has a greater energy than an electron probe for the same wavelength.

Wavelength of light emitted from the probe,

Mass of an electron, *me*=

Planck’s constant, *h*=

Charge on an electron, *e*=

The kinetic energy of the electron is given as:

s

Where, *v*= Velocity of the electron

= Momentum (*p*) of the electron

According to the de Broglie principle, the de Broglie wavelength is given as:

Energy of a photon,

Hence, a photon has a greater energy than an electron for the same wavelength.

**24.(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain.**

**(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature. Hence explain why a fast neutron beam needs to be the rmalised with the environment before it can be used for neutron diffraction experiments.**

**Ans.(a)** De Broglie wavelength =; neutron is not suitable for the diffraction experiment

Kinetic energy of the neutron, *K*= 150 eV

Mass of a neutron, *mn*=

The kinetic energy of the neutron is given by the relation:

Where, *v* = Velocity of the neutron

= Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

It is clear that wavelength is inversely proportional to the square root of mass.

Hence, wavelength decreases with increase in mass and vice versa.

It is given in the previous problem that the inter-atomic spacing of a crystal is about , i.e., . Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.

**(b)** De Broglie wavelength =

Room temperature, = 27 + 273 = 300 K

The average kinetic energy of the neutron is given as:

Where, *k *= Boltzmann constant =

The wavelength of the neutron is given as:

This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high-energy neutron beam should first be the rmalised, before using it for diffraction.

**25.An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?**

**Ans.**Electrons are accelerated by a voltage, *V*= 50 kV =

Charge on an electron, *e*=

Mass of an electron, *me*=

Wavelength of yellow light =

The kinetic energy of the electron is given as:

*E*= *eV*

De Broglie wavelength is given by the relation:

This wavelength is nearly 105 times less than the wavelength of yellow light.

The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly 105times that of an optical microscope.

**26.The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)**

**Ans.**Wavelength of a proton or a neutron,

Rest mass energy of an electron:

Planck’s constant, *h*= 6.6 × 10 – 34Js

Speed of light, *c*=

The momentum of a proton or a neutron is given as:

The relativistic relation for energy (*E*) is given as:

Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.

**27. Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.**

**Ans.**De Broglie wavelength associated with He atom =

Room temperature, *T*= = 27 + 273 = 300 K

Atmospheric pressure, *P*= 1 atm =

Atomic weight of a He atom = 4

Avogadro’s number, NA=

Boltzmann constant, *k*=

Average energy of a gas at temperature *T*, is given as:

De Broglie wavelength is given by the relation:

Where, *m*= Mass of a He atom

We have the ideal gas formula:

*PV = RT*

*PV = kNT*

Where, *V*= Volume of the gas

*N*= Number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.

**28.Compute the typical de Broglie wavelength of an electron in a metal at and compare it with the mean separation between two electrons in a metal which is given to be about.**

**[ Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This in distinguish ability has many fundamental implications which you will explore in more advanced Physics courses.]**

**Ans.**Temperature, = 27 + 273 = 300 K

Mean separation between two electrons, *r*=

De Broglie wavelength of an electron is given as:

Where, *h*= Planck’s constant =

*m*= Mass of an electron =

*k*= Boltzmann constant =

Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

**29.Answer the following questions:**

**(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e; ( – 1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?**

**(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?**

**(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?**

**(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?**

**(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:**

* ***, p = **

**But while the value of is physically significant, the value of (and therefore, the value of the phase speed ) has no physical significance. Why?**

**Ans.(a)** Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.

**(b)** The basic relations for electric field and magnetic field are

respectively

These relations include *e*(electric charge), *v*(velocity), *m*(mass), *V*(potential), *r*(radius), and *B*(magnetic field). These relations give the value of velocity of an electron as and

respectively.

It can be observed from these relations that the dynamics of an electron is determined not by *e *and *m *separately, but by the ratio e/*m*.

**(c)** At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

**(d)** The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.

**(e)** The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength is significant, but the frequency associated with an electron has no direct physical significance.

Therefore, the product (phase speed) has no physical significance.

Group speed is given as:

This quantity has a physical meaning.

Dual Nature Of Radiation And Matter pdf notes

Mohd. Sharif Qualification: B.Tech (Mechanical Engineering) [Founder of Wisdom Academy] [Aim Foundation & Free-Education.In] [Engineer By Profession | Teacher By Choice] [Blogger, YouTube Creator]