Question

It is observed that some of the spectral lines in hydrogen spectrum have wavelengths almost equal to those of the spectral lines in ${\text{He}}^{+}$ ion. Out of the following, the transitions in ${\text{He}}^{+}$ that will make this possible is-

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

Answer 1

The correct option is D $n=3$ to $n=2$

The energy difference, when electron transits from $n_i$ to $n_f$

$\Delta E=13.6Z^2[\frac{1}{(n_f{)}^2}-\frac{1}{(n_i{)}^2}]$

For the transition $6\to 4$ in ${\text{He}}^{+}$ ion,

$\Delta E=13.6\times 4\times (\frac{1}{16}-\frac{1}{36})$

$=13.6\times (\frac{1}{4}-\frac{1}{9}).......\left(1\right)$

For the transition $3\to 2$ in $\text{H}$ atom is,

$\Delta E=13.6Z^2[\frac{1}{(n_f{)}^2}-\frac{1}{(n_i{)}^2}]$

$\Delta E=13.6\times (\frac{1}{4}-\frac{1}{9}).......\left(2\right)$

From (1) and (2) we get,

$\Delta E$ in ${\text{He}}^{+}$ ion, for the transition $6\to 4=$ $\Delta E$ in $\text{He}$ atom, for the transition $3\to 2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.