Question

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:

(Assuming 1 amu = 931 MeV)

• A. $2.2$
• B. $1.1$
• C. $3.1$
• D. $22$

Answer 1

The correct option is A $2.2$

First note that $7MeV$ emission will have negligible effect on the mass of the nucleus so we can assume the man of the nucleus is comes out at $24amu.$

Momentum of emitter gamma radiation is given by

$p_g=\frac{E_r}{c}$ where Ey is the energy of emitter gamma radiation

$p_g^2=\frac{{E_r}^2}{c^2}$

Final momentum of the nucleus $P_n=m_n$

$=P_n^2={\left(mv\right)}^2=2mE$

$P_n^2=2m{E}_n$

$P_g=P_n\therefore p_n^2=p_n^2$

$\frac{E{r}^2}{c^2}=2m{E}_n.$

$E_n=\frac{E{r}^2}{2m{c}^2}$

$E_n=\frac{E{r}^2}{1862m}=\frac{7^2}{1862\times 29}=1.1\times \frac{10}{-3}Meu$

$=1.1Kev.$