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

Answer

B

Explanation

For an electron accelerated through a potential V
De Broglie wavelength `(lambda) = (12.27)/(sqrt(v))Å = (12.27 xx 10^(-10))/(sqrt(10000))`
`= 12.27 xx 10^(-12)m`