De - Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
`(|e|=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31))`

Options

A. 2.4Å
B. 0.5Å
C. 1.7Å
D. 1.2Å

Answer

C

Explanation

de-Broglie wavelength
`lamda = h/p = (h)/(mv) = (h)/(sqrt(2mV))`
`lamda = (6.6 xx 10^(-34) )/(sqrt(2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx 50))`
` = 1.7 Å`