Determine the smallest wavelength of radiation that can be emitted...

The smallest wavelength of radiation emitted by a hydrogen atom corresponds to the maximum energy transition.

This maximum energy transition occurs when an electron falls from $n = \infty$ to $n = 1$ (the series limit of the Lyman series).

Using the Rydberg formula:

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

Substituting $n_1 = 1$ and $n_2 = \infty$:

\dfrac{1}{\lambda} =R \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2} \right) =$R$

$\lambda =$\dfrac{1}{R}

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

$\lambda =$\dfrac{1}{1.097 \times 10^7} \text{ m} \approx$91.1 \times 10^{-9} \text{ m} = 91.1 \text{ nm}$

Thus, the smallest wavelength is approximately $91 \text{ nm}$.