The voltage applied to an electron microscope to produce electrons of wavelength `0.50 Å` is

Options

A. 602 V
B. 50 V
C. 138 V
D. 812 V

Answer

A

Explanation

De-Broglie wavelength is `lamda=(h)/(mv)=(h)/(sqrt(2mmE))`
But `E=eV and lamda=(h)/(sqrt(2meV)),V=(h^(2))/(2melamda^(2))`
`=((6.62xx10^(-34))^(2))/((0.5xx10^(-10))^(2)xx2xx9.1xx10^(-31)xx1.6x10^(-19))`
`implies V=601.98V~~602V`