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 `lambda=(h)/(mv)=(h)/(sqrt(2mE))`
`"But "E=ev`
`lambda=(h)/(sqrt(2meV))implies V=(h^(2))/(2melambda^(2))`
`V=((6.62 xx 10^(-34))^(2))/((0.5xx10^(-10))^(2)xx2xx9.1xx10^(-31)xx1.6xx10^(-19))`
`impliesV=601.98V=602V`