In a double slit experiment, interference is obtained from electron waves produced in an electron gun supplied with voltage V. If `lambda` 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

Options

A. `(hD)/( sqrt ( 2 me V d ^(2)))`
B. `( 2 hD)/( sqrt ( m e V d ^(2)))`
C. `( hd)/( sqrt (2 me VD ^(2)))`
D. `( hd)/( sqrt ( me VD ^(2)))`

Answer

A

Explanation

`beta = (lamda D)/( d ), lamda = (h)/(P) = ( h )/( sqrt ( 2m ( KE)))`
`=( h)/( sqrt (2 meV))`
`therefore beta = ( lamda D)/( sqrt (2 me V ) ) (1)/(d) = ( h D)/( sqrt (2 me V d ^(2)))`