Question

The transition from state n = 4 to n = 3 in a $H{e}^{+}$ ion result in ultraviolet radiation. Infrared radiation will be obtained in the transition from

• A. $n=2\to n=1$
• B. $n=3\to n=2$
• C. $n=5\to n=4$
• D. $n=8\to n=6$

Answer 1

The correct option is D

$n=8\to n=6$

In case of H-spectrum UV spectrum appears from $n_1=1$, to $n_2=2,\mathrm{3....}$
Here in case of $H{e}^{+}$ spectrum UV spectrum is from $n_1$ to $n_2=4$.

${\overline{\nu }}_{UV}=R{Z}^2(\frac{1}{3^2}-\frac{1}{4^2})R\times 4(\frac{1}{3^2}-\frac{1}{4^2})$

In case of H-spectrum IR appears from $n_1=3,n_2=4,5....$

${\overline{\nu }}_{IR}for{H}_2^{+}=R\times 1^2(\frac{1}{3^2}-\frac{1}{4^2}).....\left(i\right)$

${\overline{\nu }}_{IR}forH{e}^{+}=R\times 4(\frac{1}{n_1^2}-\frac{1}{n_2^2})=R(\frac{4}{n_1^2}-\frac{4}{n_2^2}).....\left(ii\right)$

Comparing the coefficient of equation (i) and (ii), we get
$\frac{4}{n_1^2}=\frac{1}{9};n_1=6$
$\frac{4}{n_1^2}=\frac{1}{16};n_2=8$

Thus, transition for $H{e}^{+}$ in IR is $n_1\to 6$ to $n_2\to 8$.
Note: IR region starts with $n_1=3$.