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Question Number 22813 by Tinkutara last updated on 22/Oct/17
The nucleus Fe^(57)  emits a γ-ray of energy  14.4 keV. If the mass of the nucleus is  56.935 u, calculate the recoil energy of  nucleus.
$$\mathrm{The}\:\mathrm{nucleus}\:\mathrm{Fe}^{\mathrm{57}} \:\mathrm{emits}\:\mathrm{a}\:\gamma-\mathrm{ray}\:\mathrm{of}\:\mathrm{energy} \\ $$$$\mathrm{14}.\mathrm{4}\:\mathrm{keV}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nucleus}\:\mathrm{is} \\ $$$$\mathrm{56}.\mathrm{935}\:\mathrm{u},\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{recoil}\:\mathrm{energy}\:\mathrm{of} \\ $$$$\mathrm{nucleus}. \\ $$
Answered by ajfour last updated on 23/Oct/17
p=(U/c)  K=(p^2 /(2M)) =(U^2 /(2Mc^2 ))             =(((14.4keV)^2 )/(2(56.935u)(931.5×10^3 keV/u)))  =((14.4×14.4)/(2×56.935×0.9315)) ×10^(−6) keV   ⇒   K=1.95×10^(−6)  keV .
$${p}=\frac{{U}}{{c}} \\ $$$${K}=\frac{{p}^{\mathrm{2}} }{\mathrm{2}{M}}\:=\frac{{U}^{\mathrm{2}} }{\mathrm{2}{Mc}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{14}.\mathrm{4}{keV}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{56}.\mathrm{935}{u}\right)\left(\mathrm{931}.\mathrm{5}×\mathrm{10}^{\mathrm{3}} {keV}/{u}\right)} \\ $$$$=\frac{\mathrm{14}.\mathrm{4}×\mathrm{14}.\mathrm{4}}{\mathrm{2}×\mathrm{56}.\mathrm{935}×\mathrm{0}.\mathrm{9315}}\:×\mathrm{10}^{−\mathrm{6}} {keV} \\ $$$$\:\Rightarrow\:\:\:{K}=\mathrm{1}.\mathrm{95}×\mathrm{10}^{−\mathrm{6}} \:{keV}\:. \\ $$
Commented by Tinkutara last updated on 23/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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