An isolated hydrogen atom emits a photon of `10.2 eV`.
(i) Determine the momentum of photon emitted (ii) Calculated the recoil momenium of the atom
(ii) Find the kinetic energy of the recoil atom [Mass of proton `= m_(0) = 1.67 xx 10^(-27) kg`]

Explanation

(i)Momentum of the photon is `P_1=L^2/C=(10.2xx1.6xx10^(-19))/(3xx10^8)=5.44xx10^(-27)` kg m/s
(ii)Applying the momentum conservation
`P_2=P_1=5.44xx10^(-27)` kg m/s
(iii)`k=1/2mV^2` (V=recoil speed of atom , m=mass of hydrogen atom )
`k=1/2m(P/m)^2=P^2/"2m"`
Substituting the value of the momentum of atom we get
`k=((5.44xx10^(-27))^2)/(2xx1.67xx10^(-27))=8.86xx10^(-27)` J