The basic problem is removing energy from an excited state by emission of a photon exactly equal to the energy of the transition.

So

$$ E_2 - E_1 = hf $$

If a beta decay happens such that the decay can produce beta particles of two energies, find the energy of the associated gamma photon.

So this happens as one beta decay happens to an excited state which then de-excites emiting a gamma photon to reach the ground state.

$$ E_2 - E_1 = hf $$

$$ hf = 1.015 MeV - 0.834 MeV $$

$$ hf = 0.181 MeV $$

$$ \lambda = 6.84 \times 10^{-12} \text{ m} $$