Question

An electron in a hydrogen like species is in the excited state $n_2$. The wavelength for the transition $n_2$ to $n_1=2$ is $48.24nm$. The corresponding wavelength for the transition from $n_2$ to $n_1=3$ is $142.46nm$. Find the value of $n_2$ and $Z$ and report the $H$ like atom.

• A. $n_2=5,Z=4$
• B. $n_2=4,Z=1$
• C. $n_2=7,Z=3$
• D. $n_2=4,Z=2$

Answer 1

The correct option is C $n_2=7,Z=3$

For the transition $n_2\to 2$, the wavelength is 48.24 nm or $48.24\times {10}^{-7}cm$
$\frac{1}{\lambda }=109678Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\frac{1}{48.24\times {10}^{-7}}=109678Z^2(\frac{1}{2^2}-\frac{1}{n_2^2})$
$1.89=Z^2(0.25-\frac{1}{n_2^2})$......(1)
For the transition $n_2\to 3$, the wavelength is 142.46 nm or $142.46\times {10}^{-7}cm$
$\frac{1}{\lambda }=109678Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\frac{1}{142.46\times {10}^{-7}}=109678Z^2(\frac{1}{3^2}-\frac{1}{n_2^2})$
$0.64=Z^2(0.111-\frac{1}{n_2^2})$......(2)
Divide equation (1) with equation (2)
$\frac{1.89}{0.64}=\frac{(0.25-\frac{1}{n_2^2})}{(0.111-\frac{1}{n_2^2})}$
$0.3278-\frac{2.95}{n_2^2}=0.25-\frac{1}{n_2^2}$
$0.0778=\frac{3.95}{n_2^2}$
$n_2^2=50.77$
$n_2=7$
Substitute $n_2=7$ in the equation (1)
$1.89=Z^2(0.25-\frac{1}{7^2})$
$1.89=Z^2\left(0.229\right)$
$8.2=Z^2$
$Z\approx 3$