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

Using Rydberg's 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 $n = 5$ to $n = 4$, we have $n_1 = 4$ and $n_2 = 5$.

\dfrac{1}{\lambda} =$R_H \left( \dfrac{1}{4^2} - \dfrac{1}{5^2} \right)$

\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{16} - \dfrac{1}{25} \right)

\dfrac{1}{\lambda} =R_H \left(\dfrac{25 - 16}{400} \right) = \dfrac{9 R_H}{400}

$\lambda =$\dfrac{400}{9 R_H}

Substituting the value of the Rydberg constant $R_H = 1.097 \times 10^7 \text{ m}^{-1}$:

$\lambda =$\dfrac{400}{9 \times 1.097 \times 10^7} \text{ m}

$\lambda =$\dfrac{400}{9.873 \times 10^7} \text{ m} \approx$40.51 \times 10^{-7} \text{ m}$

$\lambda \approx 4051 \text{ nm}$

The closest value among the given options is $4050 \text{ nm}$.