1. Write the decay reaction

$^{226}_{88}\mathrm{Ra} \rightarrow \, ^{222}_{86}\mathrm{Rn}^* + \alpha$

Then,

$^{222}_{86}\mathrm{Rn}^* \rightarrow \, ^{222}_{86}\mathrm{Rn} + \gamma$

Here $\mathrm{Rn}^*$ is the excited daughter nucleus.

2. Find the total decay energy using mass defect

Using the given atomic masses:

• $M(^{226}\mathrm{Ra}) = 226.005\,u$
• $M(^{222}\mathrm{Rn}) = 222.000\,u$
• $M(\alpha) = 4.000\,u$

So the mass defect is

$\Delta m = 226.005 - (222.000 + 4.000) = 0.005\,u$

Hence,

$Q = \Delta m\,c^2 = 0.005 \times 931\,\text{MeV} = 4.655\,\text{MeV}$

This is the total energy released if radon were formed in the ground state.

3. Relate alpha-particle kinetic energy and recoil energy

Since the initial Ra nucleus is at rest, after decay the $\alpha$ particle and daughter nucleus have equal and opposite momenta.

If $K_\alpha = 4.44\,\text{MeV}$, then recoil energy of daughter is

$K_R = K_\alpha \frac{m_\alpha}{m_R}$

because for same momentum, kinetic energy varies inversely with mass.

Now,

$m_\alpha = 4\,u, \qquad m_R = 222\,u$

So,

$K_R = 4.44 \times \frac{4}{222}$

$K_R = 4.44 \times 0.018018 \approx 0.080\,\text{MeV}$

4. Energy balance for decay to excited state

Let excitation energy of $^{222}\mathrm{Rn}^*$ be $E^*$.

Then,

$Q = K_\alpha + K_R + E^*$

Therefore,

$E^* = Q - K_\alpha - K_R$

Substitute values:

$E^* = 4.655 - 4.44 - 0.080$

$E^* \approx 0.135\,\text{MeV}$

Thus,

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

5. Final answer

$\boxed{135}$

The emitted $\gamma$ photon has energy $135\,\text{keV}$.