# CBSE Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter Important Question Answer

## Very Short Type Question – 1 Mark

**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

## Short type question answer – 2 Marks

**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.

## Short type question answer – 3 Marks

**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.

## Long term question answer – 5 Marks

**1. 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.

**2.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.

**3.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.

**4. 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.

**5.(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.

**6. 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.

**7. (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*=

**8.(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:Exercises11.20(b)and11.21(b)takeyoutorelativisticmechanicswhichisbeyondthescopeofthisbook.Theyhavebeeninsertedheresimplytoemphasisethepointthattheformulasyouuseinpart(a)oftheexercisesarenotvalidatveryhighspeedsorenergies.Seeanswersattheendtoknowwhat‘veryhighspeedorenergy′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:

**9.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

**10.(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.

**11.****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.

**12.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.

**13.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:

**14.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.

**15.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.

**16.(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.

**17.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

**,**

Mass of an electron,

*e*=*me*

**=**

Wavelength of yellow light

**=**

The kinetic energy of the electron is given as:

**De Broglie wavelength is given by the relation:**

*E*=*eV***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.