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

The smallest wavelength of radiation corresponds to the maximum energy transition, which occurs when an electron transitions from $n_2 = \infty$ to $n_1 = 1$.

Using the Rydberg formula:

$\dfrac{1}{\lambda} = R_H Z^2 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)<br><br>For \text{He}^+, the atomic number$Z = 2$. Substituting the values:<br><br>\dfrac{1}{\lambda} =$R_H (2)^2 \left( \dfrac{1}{1^2} - \dfrac{1}{\infty^2} \right)$<br><br>$\dfrac{1}{\lambda} = 4 R_H$<br><br>$\lambda = \dfrac{1}{4 R_H}$<br><br>We know that$\dfrac{1}{R_H} \approx 91.2 \text{ nm}$.<br><br>$\lambda = \dfrac{91.2}{4} \text{ nm} = 22.8 \text{ nm} \approx 23 \text{ nm}