Question

For the transition from $n=2\to n=1$, which of the following will produce shortest wavelength?

• A. H atom
• B. D atom
• C. ${He}^{+}ion$
• D. ${Li}^{2+}$ ion

Answer 1

The correct option is D ${Li}^{2+}$ ion

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

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