Question

The electronic transitions from $n=2$ to $n=1$ will produce shortest wavelength in (where $n$ = principal quantum state)

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

Answer 1

Answer: (A)

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)Z^2$ $=R\left(\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2}\right)Z^2$
$\frac{1}{\lambda }=\frac{3}{4}R{Z}^2$
$\therefore \lambda \propto \frac{1}{Z^2}$
$\therefore$ For shortest $\lambda ,Z$ must be maximum, which is for $L{i}^{2+}$.