InterviewSolution
Saved Bookmarks
| 1. |
The iron `Fe^(27)` nucleus emits a y-axis of energy 14.4KeV. If mass of nucleus is 56.935u, calculate the recoil energy of the nucleus. Take `1u=1.66xx10^(-27)kg.` |
|
Answer» The nuclear decay is represented as Accordinig to be Broglie hypothesis, linear momentum of photon, `p=(E)/(c)=(14.4xx1.6xx10^(-16)J)/(3xx10^(8)m//s)` `=7.68xx10^(-24)kgms^(-1)` From conservation of linear momentum, momentum of daughter nucleus `=` momentum of photon `p=7.68xx10^(-24)ms^(-1)` `:.` Recoil energy of nucleus `K=(p^(2))/(2m)=((7.68xx10^(-24))^(2))/(2xx56.935xx1.66xx10^(-27))` `=0.312xx10^(-21)J` `K=(0.312xx10^(-21))/(1.6xx10^(-16))keV` `=1.95xx10^(-6)keV` |
|