The potential difference V required for accelerating an electron to have de Broglie wavelength of `1 Å` is

Options

A. 100 V
B. 125 V
C. 150 V
D. 200 V

Answer

C

Explanation

de Broglie wavelength of an electron,
`lamda = (h)/(sqrt(2mqV))`
or, `V = (h^(2))/(2mq lamda^(2))`
`= ((6.6 xx 10^(-34))^(2))/(2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx (1 xx 10^(-10))^(2))`
`= 1.496 xx 10^(2) ~~ 150 V`