Question

Radiation coming from transition n = 2 to n = 1 of hydrogen atoms falls on helium ions in n = 1 and n = 2 states. What are the possible transitions of helium ions as they absorbs energy from the radiation?

Answer 1

The possible transition of Helium ion is to the transition n=2 from n=1.

EXPLANATION:

$E = 13.6 (\frac{-1}{4} + \frac{1}{1} ) = \frac{3}{4} \times 13.6 = 10.2 eV.$

It is good to check the possible transitions on He. n = 1 to 2 .  

$E_1 = 13.6 (\frac{-1}{4} + \frac{1}{1} )\times 4 = 40.8 eV$

[E < E1 so, it is not possible]

To n = 3 from n = 1

$E_2 =13.6 (\frac{-1}{9} +1)\times 4 = 48.3 eV$  

[E < E2 so, it is possible]

Likewise, to n = 4 from n = 1 is also impossible.

To n=3 from n = 2

 $E_3 = 13.6 (\frac{-1}{9} +\frac{1}{4}) \times 4 = 7.56 eV.$

$\begin{aligned}&n=2 \text { to } n=4\\&\mathrm{E}_{4}=4 \times 13.6(\frac{1}{4} -\frac{1}{16} )=10.2 \mathrm{eV}\\&\text { As, } \mathrm{E}_{3}<\mathrm{E} \text { and } \mathrm{E}_{4}=\mathrm{E}\end{aligned}$

Hence $E_3$ and $E_4$ are possible