An electron is accelerated through a potential difference of `10,000V`. Its de Broglie wavelength is, (nearly): `(me=9xx10^(-31)kg)`

Options

A. 12.2 nm
B. `12.2 xx10^(-13) m`
C. `12.2 xx 10^(-12)m`
D. `12.2 xx 10^(-14)m`

Answer

C

Explanation

de Broglie wavelength of electron , `lamda_(e0) = (12.27Å)/(sqrt(V("in" V)))`
Here, V = 10000 V
`:. lamda_e=(12.27)/(sqrt(1000))xx10^(-10)m = 12.27 xx 10^(-12)m`