Dual Nature of Radiation and Matter Class 12 Notes CBSE Physics Chapter 11 (Free PDF Download)

Section-A (1 Mark Questions)

1. Define photoelectric work function. How is it related to threshold frequency?

Ans. The minimum amount of radiant energy needed to pull an electron (without imparting it any kinetic energy) from a metallic surface is called work function of the metal. The relation between work function W0 and threshold frequency ν0 is W0 = hν0. 

2. Which photon has more energy: A red one or a violet one? 

Ans. Violet photon has more energy, because energy of a photon, E = hν and νviolet > νred .  

3. Why are alkali metals most suited as photo-sensitive metals?

Ans. Alkali metals have low work function. Even visible radiation can eject out electrons from them. So, alkali metals are most suitable photo-sensitive metals. 

4. Work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.

Ans. No, energy of a single photon must be greater than the work function of the metal for the emission of a photoelectron. 

5. Does the 'stopping potential' in photo-electric emission depend upon (i) the intensity of the incident radiation in a photocell? (ii) the frequency of the incident radiation?

Ans. 

(i) No, the stopping potential does not depend on the intensity of the incident radiation. 

(ii) Yes, the stopping potential depends on the frequency of incident radiation, above the threshold frequency, V0 ∝ ν. 

6. How does the maximum kinetic energy of electrons emitted vary with the work function of the metal?

Ans. The maximum kinetic energy of emitted electrons, $K_{max}=\dfrac{1}{2}mv^{2}_{max}=hv-W_{0}$ .

7. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Ans. $K_{max}=eV_{0}=1\cdot 6\times 10^{-19}C\times 1\cdot 5V$

$=24\times 10^{-19}J$ or 1.5eV

8. Why are de-Broglie waves associated with a moving football not visible? 

Ans. Due to the large mass of a football(in comparison subatomic particles), the de-Broglie wavelength $\left ( \lambda =\dfrac{h}{mv} \right )$ associated with a moving football is very small, quite beyond measurement. Hence its wave nature is not visible.

9. The de-Broglie wavelength of a particle of kinetic energy K is $\lambda$ . What would be the wavelength of the particle if its kinetic energy were K/4?

Ans. The de-Broglie wavelength of a particle of mass m and kinetic energy K is $\lambda=\dfrac{h}{\sqrt{2mK}}$ .

When the kinetic energy is $\dfrac{K}{4}$ .

${\lambda }'=\dfrac{h}{\sqrt{2m\dfrac{K}{4}}}=\dfrac{2h}{\sqrt{2mK}}=2\lambda$

10. An $\alpha -$ particle and a proton are accelerated through the same potential difference. Calculate the ratio of linear momenta acquired by the two. 

Ans. The kinetic energy gained by a particle when accelerated through potential difference V is $\dfrac{1}{2}mv^{2}=qV$ or $m^{2}v^{2}=2mqV$

∴ momentum $p=mv=\sqrt{2mqV}$

∴ $\dfrac{p\alpha }{p_{p}}=\sqrt{\dfrac{2m_{\alpha }q_{\alpha }V}{2m_{p}q_{p}V}}=\sqrt{\dfrac{2\times 4m_{p}\times 2e\times V}{2\times M_{P}\times e\times V}}=2\sqrt{2:1}$

Section-B (2 Marks Questions)

11. How is the photoelectric current affected on increasing the (i) frequency (ii) intensity of the incident radiations and why?

Ans.

(i) The increase in frequency of incident radiation has no effect on the photoelectric current. This is because the incident photon of increased energy cannot eject more than one electron from the metal surface. 

(ii) The photoelectric current increases proportionally with the increase in intensity of incident radiation. Larger the intensity of incident radiation, larger is the number of incident photons and hence larger is the number of electrons ejected from the metallic surface. 

12. What will happen to:

(i) kinetic energy of photoelectrons

(ii) photocurrent

If the light is changed from ultraviolet to X-rays in a photo-cell experiment? Intensity of the beam is the same in both cases.

Ans.

(i)As kinetic energy is proportional to the frequency of incident radiation and frequency of X-rays is higher than that of ultraviolet rays, so kinetic energy of photoelectrons will increase.

(ii) Photoelectric current will remain same as the intensity in both cases is same.

13. Two lines A and B in the plot given below show the variation of de-Broglie wavelength, $\lambda$ versus $\dfrac{1}{\sqrt{V}}$ where V is the accelerating potential difference, for two particles carrying the same charge. Which one of two represents a particle of smaller mass? 

Ans. According to de Broglie wavelength, $lambda =\dfrac{h}{\sqrt{2meV}}$

$\dfrac{\lambda }{1/\sqrt{V}}=\dfrac{h}{\sqrt{2me}}\Rightarrow \dfrac{\lambda }{1/\sqrt{V}}=\dfrac{1}{\sqrt{m}}\times \dfrac{h}{\sqrt{2e}}$

The slope is given by, $\dfrac{\lambda }{1/\sqrt{V}}=$ slope $\varpropto\dfrac{1}{\sqrt{m}}$

Slope of B > slope of A

14. The given graph shows the variation of photoelectric current (i) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. 

Ans. The pairs (2, 4) and (1, 3) have same intensity but different material.  

15. Find the ratio of de-Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V.

Ans. $\lambda \varpropto\dfrac{1}{\sqrt{V}}\dfrac{\lambda _{1}}{\lambda _{2}}=\sqrt{\dfrac{V_{2}}{V_{1}}}=\sqrt{\dfrac{36}{25}}=\dfrac{6}{5}$

$\therefore \lambda _{1}:\lambda _{2}::6:5$

PDF Summary - Class 12 Physics Dual Nature of Matter and Radiation Notes (Chapter 11)

Class 12 Physics Dual Nature of Radiation and Matter Revision Notes

Dual Nature of Radiation and Matter Class 12 Physics Revision Notes cover the following topics for your board exam.

Concise NCERT Class 12 Revision Notes Physics Chapter 11 Solution will help you to learn and understand the above-mentioned topics. Our subject matter experts have prepared the NCERT Solutions Chapter 11 Class 12 Physics Revision Notes for the perusal of all students.

Solved Sample NCERT Chapter 11 Questions Answers

Q1: The work function of a Cs (Cesium) metal is 2.14 eV. Light of frequency 6 x 1014 Hz incidents on it because of which emission of photoelectrons occurs. Find the following:

1. Maximum KE of the emitted electrons,

2. Stopping Potential, and

3. The maximum speed of emitted photoelectrons

Given: Here, Φo = 2.14 eV = 2.14 x 1.6 x 10-19J, 

f = 6 x 1014 Hz,

h (Planck’s Constant) = 6.6 x 10-34Js,

mass of an electron, me=  9.1 x 10-31 Kg

Sol:  

1.  Maximum KE =  hf - Φo = 

                              = 6.6 x 10-34Js x 6 x 1014 - 2.14 x 1.6 x 10-19J

                   Maximum KE   = 5.36 x 10-20J

 2. Stopping Potential Vo =\[\frac{K_{max}}{e}\]= (5.36 x 10-20/1.6 x 10-19)

                          Vo  = 0.335 eV

3. Maximum speed of emitted photoelectrons 

     We know that KE = ½ mv(max)2= 5.36 x 10-20J

    So, v(max)2 = (2 x 5.36 x 10-20/9.1 x 10-31)= 1.178 x 1011

or, vmax = 343 km/s

Q2: If the cut-off potential of a certain experiment is 1.5 V, then what will be the maximum KE of emitted photoelectrons?

Given:  Vo  =  1.5 V, e = 1.6 x 10-19C

To find: KEMAX =?

Solution: We know that KEMAX = e Vo 

                                                 = 1.6 x 10-19 x  1.5 

                                        KEMAX = 2.4 x 10-19J 

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