Question

Radiation com in $g$ 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

Answer: (B)

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
$E=13.6\times 2^2\times \left(\frac{1}{1^2}-\frac{1}{4^2}\right)=13.6\times \frac{15}{4}eV$
(2) n = 2 to n = 4
$E=13.6\times 2^2\times \left(\frac{1}{2^2}-\frac{1}{4^2}\right)=13.6\times \frac{3}{4}eV$
(3) n = 2 to n = 5
$E=13.6\times 2^2\times \left(\frac{1}{2^2}-\frac{1}{5^2}\right)=13.6\times \frac{21}{25}eV$
(4) n = 2 to n = 3
$E=13.6\times 2^2\times \left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\times \frac{5}{9}eV$
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 $n=1$.