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)/(lamda) = E _(5) - E _(1) " "(1)`
`lamda = ( h)/(mv) (2) therefore ("hmvc")/(h) = E _(3) - E _(1) (3)`
`E _(1) =- 13. 6 eV, E _(5) = ( - 13. 6 eV)/((5) ^(2))= - 0. 544 eV`
`m = 1 67 xx 10 ^(-27) kg and c = 3 xx 10 ^(8) ms ^(-1)`
Using these values in Eq. (3) we have
`v = (E _(5) - E _(1))/( mc )= ([ -544 + 13. 6] x (1. 6 x 10 ^(-19)))/( (1. 67 xx 10 ^(-27)) xx ( 3 xx 10 ^(-8)))= 4.17 m//s`