Question

Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on $H{e}^{+}$ ions in $n=1$ and $n=2$ states. The possible transition of helium ions as they absorb energy from the radiation is

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

Answer 1

The correct option is B $n=2\to n=4$

Energy released for transition $n=2$ to $n=1$ of hydrogen atom
$E=13.6Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$Z=1,n_1=1,n_2=2$
$E=13.6\times 1\times (\frac{1}{1^2}-\frac{1}{2^2})$
$E=13.6\times \frac{3}{4}eV$
For $H{e}^{+}$ ion, $Z=2$
(1) $n=1$ to $n=4$
$E=13.6\times 2^2\times (\frac{1}{1^2}-\frac{1}{4^2})=13.6\times \frac{15}{4}eV$
(2) $n=2$ to $n=4$
$E=13.6\times 2^2\times (\frac{1}{2^2}-\frac{1}{4^2})=13.6\times \frac{3}{4}eV$
(3) $n=2$ to $n=5$
$E=13.6\times 2^2\times (\frac{1}{2^2}-\frac{1}{5^2})=13.6\times \frac{21}{25}eV$
(4) $n=2$ to $n=3$
$E=13.6\times 2^2\times (\frac{1}{2^2}-\frac{1}{3^2})=13.6\times \frac{5}{9}eV$
Energy required for transition of $H{e}^{+}$ for $n=2$ to $n=4$ matches exactly with energy released in transition of $H$ for $n=2$ to $n=1$.