A potential difference of 15 KV is applied to accelerate the electron in an electron microscope. The de broglie wavelength of the electron waves is

Options

A. `1 Å`
B. `0.1 Å`
C. `0.5 Å`
D. `0.01 Å`

Answer

B

Explanation

`lambda=12.27/sqrt(v)=12.27/(15xx10^(3))=0.1 Å`