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

Explanation

Step I. velocity of electron.
`lambda=(h)/(mv)orv=(h)/(m lambda)`
`v=((6.62xx10^(-34)"kg "m^(2)s^(-1)))/((9.1xx10^(-31)kg)(1.54xx10^(-10)m))=4.72xx10^(6)m s^(-1)`
Step II. Potential of the electron.
`KE=1//2mv^(2)=(1)/(2)(9.1xx10^(-31)kg)(4.72xx10^(6)m s^(-1))^(2)`
`=1.01xx10^(-17) "kg "m^(2)s^(-1)=1.01xx10^(-17)J " " ( . :. 1J=1 "kg "m^(2)s^(-1))`
`=((1.01xx10^(-17)J)(1 eV))/((1.60xx10^(-19)J))=63.12 eV`