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

The correct option is A ${Li}^{2+}$

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

$=R(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2})Z^2$

$\frac{1}{\lambda }=\frac{3}{4}R{Z}^2$

$\lambda \propto \frac{1}{Z^2}$
$Z_{Li}=3,Z_{He}=2,Z_H=1$ ,

$\therefore$ For shortest $\lambda$, $Z$ must be maximum which is for ${Li}^{2+}$.

Option A is Correct