An atom recoils when it emits a photon. The energy of the photon...

The energy difference between the states for a transition from $n = 3$ to $n = 2$ in a hydrogen atom is given by:

$\Delta E = 13.6 \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) \text{ eV} = 13.6 \times \dfrac{5}{36} \text{ eV} \approx 1.89 \text{ eV}$

Let $E$ be the energy of the emitted photon and $K$ be the recoil kinetic energy of the atom. From the conservation of energy:

$\Delta E = E + K$

From the conservation of momentum, the momentum of the recoiling atom equals the momentum of the photon:

$p = \dfrac{E}{c}$

The recoil kinetic energy is:

$K = \dfrac{p^2}{2m} = \dfrac{E^2}{2mc^2} \approx \dfrac{(\Delta E)^2}{2mc^2}$

The energy of the emitted photon is $E = \Delta E - K$. The wavelength of the emitted photon is $\lambda = \dfrac{hc}{E}$.

\lambda = \dfrac{hc}{\Delta E - K} = \dfrac{hc}{\Delta E} \left( 1 -\dfrac{K}{\Delta E} \right)^{-1} \approx \dfrac{hc}{\Delta E} \left($1 +$\dfrac{K}{\Delta E} \right)

The fractional change in wavelength is:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{K}{\Delta E} = \dfrac{\Delta E}{2mc^2}

Using the rest mass energy of a hydrogen atom $mc^2 \approx 938 \times 10^6 \text{ eV}$:

\dfrac{\Delta \lambda}{\lambda} = \dfrac{1.89}{2 \times 938 \times 10^6} \approx$1 \times 10^{-9}$