The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearly

Options

A. `100 Å`
B. `123 Å`
C. `1.23 Å`
D. `0.123 Å`

Answer

C

Explanation

`lambda = (1)/(sqrt( 2 m Q V))`
` = (6.6 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx 100))`
`= 1.23 Å`