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Q.
The transition of the electron takes place from $n$ = 2 orbit to $n$ = 1 orbit. Which of the following gives the shortest wavelength ?
Atoms
Solution:
According to Rydberg equation:
$\frac{1}{\lambda} = RZ^{2}\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)$
So, keeping $R, n_1, n_2$ same, wavelength is inversely proportional to the square of atomic number $(Z)$.
For $H$-atom, $Z = 1$
For $D$ atom, $Z = 1$
For $He^+$ ion, $Z = 2$
For $Li^{2+}$ ion, $Z = 3$
So, shortest wavelength of transition will correspond to maximum $Z^{2+}$ i.e. for $Li^{2+}$ ion