If the shortest wavelength of the Lyman series for a hydrogen atom...

The Rydberg formula for the wavelength of emitted radiation is given by:

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

For the shortest wavelength of the Lyman series for a hydrogen atom, $Z = 1$, $n_1 = 1$, and $n_2 = \infty$. Substituting these values:

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

For the longest wavelength of the Paschen series for a $\mathrm{He}^{+}$ ion, $Z = 2$, $n_1 = 3$, and $n_2 = 4$. Let this wavelength be $\lambda$. Substituting these values:

$\dfrac{1}{\lambda} = R (2)^2 \left( \dfrac{1}{3^2} - \dfrac{1}{4^2} \right)$

$\dfrac{1}{\lambda} = 4R \left( \dfrac{1}{9} - \dfrac{1}{16} \right)$

$\dfrac{1}{\lambda} = 4R \left( \dfrac{16 - 9}{144} \right) = 4R \left( \dfrac{7}{144} \right) = \dfrac{7R}{36}$

Substituting $R = \dfrac{1}{x}$ into the equation:

$\dfrac{1}{\lambda} = \dfrac{7}{36x}$

$\lambda = \dfrac{36x}{7}$

Answer: $\dfrac{36x}{7}$