If the total energy of an electron in a hydrogen atom in excited state is - 3.4 eV, then the de Broglie wavelength of the electron is

Options

A. `6.6xx10^(-12)` m
B. `3xx10^(-10)` m
C. `6.6xx10^(-10)` m
D. `9.3xx10^(-12)` m

Answer

C

Explanation

`E_n=E_1/n^2 rArr -3.4 eV = -13.6/n^2 rArr n^2= 13.6/3.4 =4`
`therefore` n=2
Velocity of electron in n =2 is
`v=(2.19xx10^6)/2 m s^(-1)`
`therefore lambda=h/(mv)=(6.6xx10^(-34)xx2)/(9.1xx10^(-31)xx2.19xx10^6)=6.6xx10^(-10)` m