Class 12 Physics Important Chapter 11 Dual Nature of Radiation and Matter

Class 12 Physics Important Chapter 11 Dual Nature of Radiation and Matter Solutions English Medium As Per The New Syllabus to each chapter is provided in the list so that you can easily browse through different chapters ASSEB Class 12 Physics Additional Solutions in English and select need one. NCERT Class 12 Physics Additional Solutions Download PDF. HS 2nd Year Physics Important Solutions.

Class 12 Physics Important Chapter 11 Dual Nature of Radiation and Matter

Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. CBSE Class 12 Physics Additional Question Answer are part of All Subject Solutions. Here we have given HS 2nd Physics Important Solutions English Medium for All Chapters, You can practice these here.

Dual Nature of Radiation and Matter

Chapter: 11

IMPORTANT QUESTION AND ANSWER

Answer The Following Questions:

1. If the maximum kinetic energy of electrons emitted in a photo cell is 5 eV, what is the stopping potential?

Ans. Stopping potential:

V_o = K_max /e = 5eV /e = 5 V.

2. If the intensity of incident radiation on a metal is doubled, what happens to the kinetic energy of electrons emitted?

Ans: Kinetic energy of photo-electrons remains unchanged. It does not depend on the intensity of incident radiation.

3. How does the stopping potential applied to a photo cell change, if the distance between the light source and the cathode of the cell is doubled?

Ans. The stopping potential remains same. When the distance is doubled, the intensity of incident radiation becomes one fourth of the original intensity. But the stopping potential is independent of intensity.

4. What happens to the wavelength of a photon after it collides with an electron.

Ans. A photon transfers a part of its energy to the colliding electron, so its energy decreases and consequently wavelength increases. 

5. Write Einstein’s photoelectric equation.

Ans: According to Einstein’s photoelectric equation, the maximum kinetic energy of a photoelectric is given by:

K= 1/2 mv² _max = hv-hv₀ 

where v = frequency of incident radiation, and W₀ = work function of the metal.

6. Why no electron is emitted from a plastic plate, when light from a bulb falls on it?

Ans. The work function of plastic is quite large. The energy of the photons of visible light is less than W₀ so there is no electron emission.

7. Light from the bulb falls on the wooden table, but no photo-electrons are emitted, why?

Ans: The work function for the wood is much higher than the energy of the photon of visible light coming from bulb.

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

Ans. The minimum amount of radiant energy needed to pull an electron (as that 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 v₀ is W₀ = hv₀

9. On what factor does the stopping potential depend?

Ans. It depends upon the frequency of incident radiation or kinetic energy of photoelectron.

10. Find the dimension of Planck’s constant.

Ans: Planck’s constant hhh appears in the equation E=hν where E is energy and ν\nuν is frequency. Since energy has the dimensions of mass, length, and time ([ML²T⁻²] and frequency has the dimension of [T⁻¹], the dimensions of hhh are:

[h ]= [ML²T⁻¹]

Thus, the dimension of Planck’s constant is [ML²T⁻¹]

11. What is the dimension of work function?

Ans: The work function ϕ0​ represents the minimum energy required to remove an electron from the surface of a metal. 

Its dimensions are the same as energy:

[ϕ0​] = [ML²T⁻²]

Thus, the dimension of work function is [ML²T⁻²].