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

Options

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

Answer

C

Explanation

de-Broglie wavelength of electron `(lamda_e)=(12.27)/sqrtVÅ`
V = accelerating voltage
`lamda_e=(12.27)/(10000)xx10^(-10)m`
`lamda_e=12.2 xx10^(-12)m `