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In a double slit experiment, interference is obtained from electron waves produced in an electron gun supplied with voltage V. If l is wavelength of the beam, D is the distance of screen, d is the spacing between coherent source, h is Planck’s constant, e is charge on electron and m is mass of electron, then fringe width is given as

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$\frac{h D}{\sqrt{2 m e V} d}$

$\frac{2 h D}{\sqrt{m e V} d}$

$\frac{h d}{\sqrt{2 m e V} D}$

$\frac{2 h d}{\sqrt{2 m e V} D}$

answer is A.

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Detailed Solution

in Young's double slit experiment the fringe width is $β = \frac{λ D}{d}$

according to de Broglie wavelength of the electron is $λ = \frac{h}{p} w h e r e k i n e t i c e n e r g y o f t h e e l e c t r o n \frac{p^{2}}{2 m} = e V a n d p = \sqrt{2 m e V}$

so $β = \frac{h}{\sqrt{2 m e V}} \frac{D}{d}$

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