An electron beam can undergo defraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes `1.54` Å?

Answer

C

Explanation

For `1.54 Å` wavelength, the momentum for electron would be given by the expression
`p = (h)/(lambda) rArr v = (h)/(m lambda)`
The potential to which a beam of electron be acceletated is given by the expression
`eV = (1)/(mv^(2))`
`i.e. V = (1)/(2me) ((h)/(lambda))^(2)`
Substituting the value , we get
`V = (1)/((2)(9.1 xx 10^(-31)kg) (1.6 xx 10^(-19)C))((6.626 xx 10^(-34) J s)/(1.54 xx 10^(-10)m))^(2)`
`= 63.573 m s^(-1)`