Determine the wavelength of the radiation emitted by hydrogen...

Using the Rydberg formula for the wavelength of emitted radiation:

\dfrac{1}{\lambda} = R_{H} \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

For the transition from $n = 3$ to $n = 2$, we have $n_1 = 2$ and $n_2 = 3$.

\dfrac{1}{\lambda} =1.097 \times 10^7 \left(\dfrac{1}{2^2} - \dfrac{1}{3^2} \right) \text{ m}^{-1}

\dfrac{1}{\lambda} =1.097 \times 10^7 \left(\dfrac{1}{4} - \dfrac{1}{9} \right) \text{ m}^{-1}

\dfrac{1}{\lambda} =1.097 \times 10^7 \left(\dfrac{5}{36} \right) \text{ m}^{-1}

$\lambda =$\dfrac{36}{5 \times 1.097 \times 10^7} \text{ m}

$\lambda \approx 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm}$