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

Options

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

Answer

C

Explanation

de-Broglie wavelength ,`lambda=(h)/(sqrt(2meV))`
`therefore lambda=(6.6xx10^(-34))/(sqrt(2xx9)xx10^(-31)xx1.6xx10^(-19)xx10^(4))`
`=(6.6xx10^(-34+23))/(sqrt(28.8))=(6.6xx10^(-11))/(5.366)`
`=1.2299xx10^(-11)`
`therefore lambda~~12.2xx10^(-12m)`