Question

The transition of the electron takes place from $n$ = 2 orbit to $n$ = 1 orbit. Which of the following gives the shortest wavelength ?

• A. H atom
• B. D atom
• C. $H{e}^{+}$ ion
• D. $L{i}^{2+}$ ion

Answer 1

Answer: (D)

According to Rydberg equation:
$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
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 $H{e}^{+}$ ion, $Z=2$
For $L{i}^{2+}$ ion, $Z=3$
So, shortest wavelength of transition will correspond to maximum $Z^{2+}$ i.e. for $L{i}^{2+}$ ion