If the total energy of anelectron in a hydrogen like atom in excited state is `-3.4eV`,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))`
`-3.4eV=-13.6/(n^(2))` or `n^(2)=13.6/3.4=4`,
`:.n=2`
Velocity of elecron isn=2 is `(v_(1))/2`
`=(2.19xx10^(6))/2ms^(-1)`
`:.lamda=h/(mv)=(6.6xx10^(-34)xx2)/(9.1xx10^(-31)xx2.19xx10^(6))`
`=6.6xx10^(-10)m`