CBSE Class 12 Physics Modern Physics Assignment

CBSE Class 12 Physics Modern Physics Assignment. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Comprehension-1

According to Maxwell’s theory of electromagnetism, an accelerated charge particle loses its energy in the form of electromagnetic radiation. This concept is used in production of X-rays. where electrons are accelerated with the help of accelerating potential. When electrons collide with the target plate, they lose the energy. Minimum wavelength of X-ray coorresponds to maximum energy of electron and hence  hc /λ min = eV . Where V is accelerating potential. Now answer the following questions.
1. Generation of X-ray is an illustration of the
(a) Phenomenon of conversion of K.E. into radiant energy
(b) Principle of conservation of mass
(c) Principle of conservation of momentum
(d) Phenomenon of conversion of matter into energy

2. An X-ray machine has an accelerating potential difference of 25,000 volt. By calculation the shortest wavelength will be obtained as : (h = 6.62 × 10^–34 J-sec., e = 1.6 × 10^–19 coulomb)
(a) 0.25 Å (b) 0.50 Å (c) 1.00 Å (d) 2.50 Å

3. The X-ray beam coming from an X-ray tube will be
(a) monochromatic
(b) having all wavelengths smaller than a certain maximum wavelength
(c) having all wavelengths larger than a certain minimum wavelength
(d) having all wavelengths lying between a minimum and a maximum wavelength.

4. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation
(a) the minimum wavelenght increases (b) the minimum wavelength decreases
(c) the intensity remains decreases (d) the minimum wavelength is unchanged

Comprehension-2

Incase of photoelectric effect, we know the famous Einstiens formula of Kmax = hv – φ. Where φ is work function of metal from which photoelectrons has to be ejected out, v is frequency of light and Kmax is maximum kinetic energy of photoelectrons. When we make plate potential sufficiently negative (stoping potential v0) current becomes zero. Which implies that K_max = eV_0. ∴ V0 = (h/e)v – (φ/e)
Now answer the following questions

5. Figure shows the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. The ratio this h/e is

(a) 10^–15 V_s (b) 2 × 10^–15 V_sV₀ (in volt) 21.656112 3 4 5_V(x 10¹⁴ Hz) (c) 3 × 10^–15 V_s (d) 4.14 × 10^–15 V_s

6. In this question, the work function is
(a) 0.212 eV (b) 0.313 eV (c) 0.414 eV (d) 0.515 eV

7. In this question cut off frequency (minimum frequency at which photoelectrons get emitted) is
(a) 1 × 10¹⁴ Hz (b) 1.656 × 10¹⁴ Hz (c) 5 × 10¹⁴ Hz (d) none of these

Each of the questions given below consists of two statements, an assertion (A) and reason (R).
Select the number corresponding to the appropriate alternative as follows
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

8. A: When an α-particle and proton are accelerated through the same potential difference, then the linear momenta acquired by two is same.
R: Linear momenta acquired by the particles = √2q₁ V, where V is the potential difference applied.

9. A: If the voltage applied to the electrodes of X-ray tube be V, then X-rays of one wave length = hc/eV are emitted.
R: Electron gives out photo of minimum energy eV.

10. A: If the frequency of incident radiation on a metal is doubled, then K.E. of photo-electrons emitted will be doubled.
R: K.E. = (1/2) mv2 = eVs/2, where v is the velocity of photo electrons and Vs is stopping potential.

11. A: The graph of stopping potential versus frequency of incident radiation is a straight line passing through the origin if threshold frequency of photocathode is zero.
R: Stopping potential V_s =hv/e − φ0/e  where φ0 is the work function.

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