Question

A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statement is correct:
A) Diffraction pattern is not observed on the screen in the case of electrons.
B) The angular width of the central maximum of the diffraction pattern will increase.
C) The angular width of the central maximum will decrease.
D) The angular width of the central maximum will remain unaffected.

Answer 1

Hint:Fluorescent screen is the transparent screen which is coated with fluorescent material so as to show the x-rays images. When a light falls upon a narrow slit pattern then there is formation of differentiation patterns.

Formula used:
The formula of the De Broglie’s equation is equal to,
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$
Where wavelength is$\lambda $, the Planck’s constant is h, mass is m and velocity is v.
The formula of the angular width is given by,
$ \Rightarrow \theta = \dfrac{\lambda }{d}$
Where the slit length is d, the wavelength is $\lambda $ and the angular width is$\theta $.

Complete step by step answer:
It is given on the problem that a parallel beam of fast moving electrons is incident normally on a narrow slit a fluorescent screen is placed at a large distance from the slit and we need to find the correct option for this problem.

Let us consider each option and then find the answer for this problem.

Option A,
Diffraction pattern is not observed on the screen in the case of electrons.
The option A is not the correct answer for this problem because we observe diffraction patterns. If a fast electron falls upon a narrow slit then a diffraction pattern is observed.

Option B,
The angular width of the central maximum of the diffraction pattern will increase.
The formula of the De Broglie’s equation is equal to,
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$
Where wavelength is $\lambda $, the Planck’s constant is h, mass is m and velocity is v.
As the wavelength of the wave is inversely proportional to the velocity of the wave and as the speed of the wave is increased and therefore the wavelength decreases. Hence option B is not the correct answer.

Option C,
The angular width of the central maximum will decrease.
The formula of the De Broglie’s equation is equal to,
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$
Where wavelength is $\lambda $, the Planck’s constant is h, mass is m and velocity is v.
Here by De Broglie’s equation it is clear that the wavelength is inversely proportional to the velocity of the wave.
As the speed of the electron is increased therefore the wavelength of the wave gets decreased.

Option D,
The angular width of the central maximum will remain unaffected.
The formula of the angular width is given by,
$ \Rightarrow \theta = \dfrac{\lambda }{d}$
Where the slit length is d, the wavelength is $\lambda $ and the angular width is$\theta $.
Now as the wavelength of the wave decreases as the velocity increases therefore the angular width will also decrease as the angular width is directly proportional to the wavelength and therefore this option is not correct.

The correct answer for this problem is option C.

Note:The differentiation pattern is the formation of dark and light fringes on the surface of the screen which is formed due to the interference of the light waves coming out from the narrow slits. The differentiation pattern due single slit consists of a central bright band and has alternate dark and light fringes.