A free nucleus of mass 24 , amu emits a gamma | vedclass

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(B) The momentum of the photon is given by $p = \frac{E}{c}$,where $E = 7 \, MeV = 7 	imes 10^6 \, eV$.By the law of conservation of momentum,the rec

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(B) The momentum of the photon is given by $p = \frac{E}{c}$,where $E = 7 \, MeV = 7 	imes 10^6 \, eV$.By the law of conservation of momentum,the recoil momentum of the nucleus $p_N$ must be equal to the momentum of the photon $p_{Ph}$.$p_N = p_{Ph} = \frac{E}{c}$.The recoil kinetic energy of the nucleus is $K = \frac{p_N^2}{2M}$.Substituting $p_N = \frac{E}{c}$,we get $K = \frac{E^2}{2Mc^2}$.Given $M = 24 \, amu$. Since $1 \, amu \approx 931.5 \, MeV/c^2$,we have $Mc^2 = 24 	imes 931.5 \, MeV = 22356 \, MeV$.Now,$K = \frac{(7 \, MeV)^2}{2 	imes 22356 \, MeV} = \frac{49}{44712} \, MeV$.$K \approx 0.0010959 \, MeV$.Converting to $keV$ by multiplying by $1000$: $K \approx 1.0959 \, keV \approx 1.1 \, keV$.

$A$ free nucleus of mass $24 \, amu$ emits a gamma photon (when initially at rest). The energy of the photon is $7 \, MeV$. The recoil energy of the nucleus in $keV$ is:

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