The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearly

Options

A. 0.123 nm
B. 0.133 nm
C. 0.143 nm
D. 0.153 nm

Answer

A

Explanation

Accelerating potential, V = 100 V
The de Broglie wavelength `lamda` is given by
`lamda = h/p = (1.227)/sqrtV` nm
`lamda = (1.227 )/(sqrt 100) nm = 0.123 nm`