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CBSE 12th Standard Physics Subject Dual Nature of Radiation and Matter Ncert Exemplar 1 Mark Questions 2021

Published on: 23/05/2021

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Questions:

Generic 2

5 x 2 = 10

Show graphically the variation of de-Broglie wavelength $\lambda $ with the potential V through which an electron is accelerated from rest.
To plot the graph between the two quantities first of all we have to find the relation between the two through the connecting formula.

A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?

Write the expression for the de-Broglie wavelength associated with a charged particle having charge q and mass m, when it is accelerated by a potential V.

For a given photosensitive material and with a source of constant frquency of incident radiation, how does the photocurrent vary with the intensity of incident light?

A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1nm. The first maximum of intensity in the reflected beam occurs at $\theta =30°$ . What is the kinetic energy E of the beam in eV?

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Answers Key:

Generic 2

5 x 2 = 10

$\lambda=\frac{1}{\sqrt{V}}$
$ \lambda^2 V=\text { constant }\\ \text {constant }=\frac{1}{\sqrt{2 m e}} $

$ \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}} \\ \lambda \propto \frac{1}{\sqrt{m}} $
electrons have smaller mass $\lambda_e>\lambda_p$. For given kinetic energy electrons have greater de-Broglie wavelength as these have smaller mass. $λ$

de-Broglie wavelength, $\lambda=\frac{b}{p}=\frac{b}{\sqrt{2 m q V}}$

The photocurrent increase linearly with the intensity of incident radiation.

$K E=\frac{1}{2} \frac{p^2}{m}=\frac{1}{2} \times \frac{\left(6.63 \times 10^{-24}\right)^2}{1.67 \times 10^{-27}} J=\frac{1}{2} \frac{\left(6.63 \times 10^{-24}\right)^2}{1.67 \times 10^{-27} \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}=0.082 \mathrm{eV}$