Question

In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?

• A. Hydrogen atom
• B. Deuterium atom
• C. Singly ionized helium
• D. Doubly ionized lithium

Answer 1

The correct option is D

Doubly ionized lithium

So the Question is asking us to find the minimum wavelength for $n_1=2$ to $n_2=1$ electron transition.
Well, since $n_1$ and $n_2$ is fixed then in equation there is only one variable that remains.
Z Atomic Number
$\frac{1}{\lambda }=R{z}^2(\frac{1}{1^2}-\frac{1}{2^2})$
$\frac{1}{\lambda }=\frac{R\left(z^2\right)3}{4}$
$\lambda =\frac{4}{R\left(z^2\right)3}$
So wavelength will be minimum for maximum z
$Z_H=1$
$Z_{Deoterium}$ = 1
$Z_{He}^{+}$ = 2
$Z_{Li}^{++}$ = 3
Hence option (d)
Doubly ionized Lithium