Question

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

• A. $Li^{+2}$
• B. $He^{+}$
• 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 $Li ^{2+}$.