InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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`] |
|
Answer» (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 |
|