Question

The transition in $H{e}^{+}$ ion that will give rise to a spectral line having the same wavelength as that of some spectral line in hydrogen atom is

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

Answer 1

Answer: (C)

In case of $H{e}^{+}$ ion the Rydberg constant happens to be four times that in case of hydogen atom. Therefore, to have the same wavelength of some spectral line one must have $\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]=4\left[\frac{1}{p_f^2}-\frac{1}{p_i^2}\right]$
where $n_f$ and $n_i$ are principal quantum numbers of final and initial orbits for hydrogen atom and $p_f$ and $p_i$ are those for $H{e}^{+}$ ion. This gives $p_f=2n_f$ and $p_i=2n_i$