The recoil speed of a hydrogen atom after it emits a photon is going form n=5 state to n =1 state is ….. m/s.

Answer

c

Explanation

For photon emitted from hydrogen atom, the wavelength is
`(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` (i)
But according to de Broglie concept
`lambda = (h)/(p) implies (1)/(lambda) = (p)/(h)` (ii)
From (i) and (ii)
`(p)/(n) = R((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) implies p = Rh ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` (iii)
Since the momentum of the hydrogen atom initially was zero . therefore finally the momentum of photon is equal to momentum of the hydrogen atom in magnitude (By law of conservation of momentum ). Let the momentum of hydrogen atom be `m_(H) V_(H)`. Then, from (iii)
`m_(H) V_(H) = R_(h) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
implies `V_(H) = (Rh)/(m_(H)) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`= (1.097 xx 10^(7) xx 6.63 xx 10^(-34))/(1.67 xx 10^(-27)) ((1)/(1^(2)) - (1)/(5^(2)))`
`implies V_(H) = 4.178 m s^(-1)`