An isolated hydrogen atom emits a photon of `10.2 eV`.
(i) Determine the momentum of photon emitted (ii) Calculate the recoil momentum of the atom
(iii) Find the kinetic energy of the recoil atom [Mass of proton `= m_(p) = 1.67 xx 10^(-27) kg`]

Explanation

(i) Momentum of the photon is `p_(1) = (E)/(c) = (10.2 xx 1.6 xx 10^(-19))/(3 xx 10^(3)) = 5.44 xx 10^(-27) kg-m//s`
(ii) Applying the momentum conservation `p_(2) = p_(1) = 5.44 xx 10^(-27) kg-m//s`

(iii) `K = 1/2 mv^(2)` (`v =` recoil speed of aotm, m = mass of hydrogen atom) `K = 1/2 m (p/m)^(2) = (p^(2))/(2m)`
Substituting the value of the momentum of atom, we get `K = ((5.44 xx 10^(-27))^(2))/(2 xx 1.67 xx 10^(-27)) = 8.86 xx 10^(-27) J`