The de Broglie wavelength of an electron in a metal at `27^(@)C` is
`("Given"m_(e)=9.1xx10^(-31)kg,k_(B)=1.38xx10^(-23)JK^(-1))`

Options

A. `6.2xx10^(-9)m`
B. `6.2xx10^(-10)m`
C. `6.2xx10^(-8)m`
D. `6.2xx10^(-7)m`

Answer

A

Explanation

(a) : Here, T=27+273=300K
For an electron in a metal, momentum `p=sqrt(3mk_(B)T)`
de Broglie wavelength of an electron is
`lamda=(h)/(p)=sqrt(3mk_(B)T)`
`=(h)/sqrt(3xx(9.1xx10^(-31))xx(1.38xx10^(-23))xx300)`
`=6.2xx10^(-9)m`