Question

A stationary nucleus of mass $24amu$ emits a gamma photon. The energy of the emitted photon is $7MeV$. The recoil energy of the nucleus is:

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

Answer 1

The correct option is B $1.1keV$

The energy of emitted photon, $E=hf=7MeV$

If $p$ be the momentum of photon and $v$ be the recoil velocity of nucleus, then by conservation of momentum

$p=mv$ or $E/c=mv$ or $v=E/mc$

Thus, recoil energy $K=\frac{1}{2}m{v}^2=\frac{m}{2}\times \frac{E^2}{m^2{c}^2}=\frac{E^2}{2m{c}^2}=\frac{(7MeV{)}^2}{2\times (24\times 931.5MeV)}=1.09\times {10}^{-3}MeV=1.1keV$

where $(1amu=931.5meV)$