The recoil speed of a hydrogen atom after it emits a photon in going from `n=5 " state to " n=1` state is

Options

A. 2.18 m/s
B. 3.36 m/s
C. 4.17 m/s
D. 5.26 m/s

Answer

C

Explanation

`hv=E_5-E_1`
`(hc)/lambda =E_5-E_1` (1)
`lambda = h/(mv) (2) :. (hmvc)/h = E_5-E_1` (3)
`E_1=-13.6eV, E_5 =(-13.6eV)/((5)^2)=-0.544 eV`
`m=1.67xx10^(-27)kg and c=3xx10^8 ms^(-1)`
Using these values in Eq. (3), we have
`v =(E_5-E_1)/(mc) =([-.544 +13.6]xx(1.6x10^(-19)))/((1.67xx10^(-27)) xx(3xx10^8)) = 4.17 m//s`
The negative sign shows the hydrogen recoils after emitting the photon.