Question

The de Broglie wavelength `lamda` of an electron accelerated through a potential V in volts is
A. `(1.227)/(sqrt(V))nm`
B. `(0.1227)/(sqrt(V))nm`
C. `(0.01227)/(sqrt(V))nm`
D. `(12.27)/(sqrt(V))nm`

Answer 1

Correct Answer - A
(a) : Consider an electron of mass m and charge e accelerated from rest through potential V. Then, K=eV
`K=(1)/(2)mv^(2)=(p^(2))/(2m):.p=sqrt(2mK)=sqrt(2meV)`
The de Broglie wavelength `lamda` of the electron is
`lamda=(h)/(p)=(h)/(sqrt(2mK))=(h)/(sqrt(2meV))`
Substituting the numerical values of h, m, e, we
`lamda=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xxV))`
`=(1.227xx10^(-9))/(sqrt(V))m=(1.227)/(sqrt(V))nm`