The de Broglie wavelength `lamda` of an electron accelerated through a potential V in volts is

Options

A. `(1.227 )/sqrtVnm`
B. `(0.227 )/sqrtVnm`
C. `(0.01227 )/sqrtVnm`
D. `(0.1227)/(sqrtV)Å`

Answer

A

Explanation

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 get
`lamda = ( 6.63 xx10^(-34))/(sqrt(2.9.1 xx10^(-31) xx1.6 xx 10^(-19)) xxV`
`= (1.227 xx10^(-9))/sqrtV m = (1.227)/sqrt(V)` nm