Energy of Gamma Photon Emitted in Alpha Decay of Radium

The total energy released ($Q_{\text{total}}$) in the decay of a $^{226}_{88}\text{Ra}$ nucleus from its ground state to the ground state of $^{222}_{86}\text{Rn}$ is given by the mass defect:

$\Delta m = M(\text{Ra}) - M(\text{Rn}) - M(\alpha)$

Given values: $M(\text{Ra}) = 226.005\text{ u}$ $M(\text{Rn}) = 222.000\text{ u}$ $M(\alpha) = 4.000\text{ u}$

$\Delta m = 226.005 - 222.000 - 4.000 = 0.005\text{ u}$

Using $1\text{ u} = 931\text{ MeV}/c^2$, the total $Q$-value is: $Q_{\text{total}} = 0.005 \times 931\text{ MeV} = 4.655\text{ MeV}$

Now, consider the $\alpha$-decay of $^{226}_{88}\text{Ra}$ at rest to an excited state of $^{222}_{86}\text{Rn}$. Let $Q_{\alpha}$ be the energy released in this step. By conservation of linear momentum, the momenta of the excited $^{222}_{86}\text{Rn}$ nucleus and the $\alpha$-particle are equal in magnitude:

$p_{\text{Rn}} = p_{\alpha}$

The kinetic energy of the recoil $^{222}_{86}\text{Rn}$ nucleus is: $K_{\text{Rn}} = \frac{p_{\text{Rn}}^2}{2 M_{\text{Rn}}} = \frac{M_{\alpha}}{M_{\text{Rn}}} K_{\alpha}$

Thus, the energy $Q_{\alpha}$ shared between the kinetic energies of the $\alpha$-particle and the recoil nucleus is: $Q_{\alpha} = K_{\alpha} + K_{\text{Rn}} = K_{\alpha} \left(1 + \frac{M_{\alpha}}{M_{\text{Rn}}}\right)$

Given $K_{\alpha} = 4.44\text{ MeV}$, $M_{\alpha} = 4.000\text{ u}$, and $M_{\text{Rn}} = 222.000\text{ u}$: $Q_{\alpha} = 4.44 \times \left(1 + \frac{4}{222}\right) = 4.44 \times \left(1 + \frac{2}{111}\right) = 4.44 \times \frac{113}{111} = 4.52\text{ MeV}$

The excitation energy $E^*$ of the $^{222}_{86}\text{Rn}$ nucleus is the difference between the total energy available and the energy released in the $\alpha$-decay:

$E^* = Q_{\text{total}} - Q_{\alpha} = 4.655\text{ MeV} - 4.520\text{ MeV} = 0.135\text{ MeV}$

When the excited $^{222}_{86}\text{Rn}$ nucleus transitions to its ground state, it emits a $\gamma$-photon of energy equal to this excitation energy (since the nuclear recoil during photon emission is negligible):

$E_{\gamma} = E^* = 0.135\text{ MeV} = 135\text{ keV}$

Answer: 135