Question

If on transition to the ground state an $He^{+}$ ion emits two photons in succession, having wavelengths $1026.7\overset{^\circ }{A}$ and $304\overset{^\circ }{A}$ , then the quantum number $n$ corresponding to the exciting state of $He^{+}$ ion is [ $R=1.096\times 10^{7}m^{- 1}$ ]

• A. 4
• B. 6
• C. 2
• D. 1

Answer 1

Answer: (B)

$\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2}=\operatorname{Rch} Z^2\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$
Put $\lambda _{1}=1026.7Å$ and $\lambda _{2}=304 \, Å$
$Z=2$ for $He^{+}$ ion
On solving for $n$
$n=6$