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?

Explanation

The aim is to find the energy of the electron in electron volts (eV). As K.E. `= (1)/(2) mv^(2)`, therefore, first we calculate velocity of the electron to have a wavelength of `1.54 Å`. Applying de Broglie equation
`lamda = (h)/(mv)`
`v = (h)/(m xx lamda) = ((6.62 xx 10^(-34) kg m^(2) s^(-1)))/((9.1 xx 10^(-31) kg) (1.54 xx 10^(-10) m)) = 4.72 xx 10^(6) m s^(-1)`
Now, K.E. `= (1)/(2) mv^(2) = (1)/(2) (9.1 xx 10^(-31)kg) (4.72 xx 10^(6) ms^(-1))^(2) = 1.01 xx 10^(-17) kg m^(2) s^(-2)`
`= 1.01 xx 10^(-17) J " " ( :' 1J = 1 kg m^(2) s^(-2))`
`= (1.01 xx 10^(-17))/(1.602 xx 10^(-19)) eV " " ( :' 1 eV = 1.602 xx 10^(-19) J)`
`= 63.12 eV`
Hence, potential required = 63.12 volts