For which of the following systems is the wavelength associated...

The wavelength of a spectral line for a hydrogen-like species is given by the Rydberg formula:

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

For the transition from $n = 2$ to $n = 1$, $n_{1} = 1$ and $n_{2} = 2$. Thus, the formula becomes:

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

From this relation, it is clear that $\lambda \propto$\dfrac{1}{Z^{2}}.

To obtain the minimum wavelength, the atomic number $Z$ must be maximum.

For doubly ionized lithium ($Li^{2+}$), $Z = 3$.
For singly ionized helium ($He^{+}$), $Z = 2$.
For deuterium atom ($D$), $Z = 1$.
For hydrogen atom ($H$), $Z = 1$.

Since doubly ionized lithium has the maximum value of $Z$, the wavelength associated with the transition from $n = 2$ to $n = 1$ is minimum for it.