Radiation coming from transitions n=2 to n=1 of hydrogen atoms fall on

Question:

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

  1. (1) $n=2 \rightarrow n=3$

  2. (2) $n=1 \rightarrow n=4$

  3. (3) $n=2 \rightarrow n=5$

  4. (4) $n=2 \rightarrow n=4$


Correct Option: , 4

Solution:

(4) Energy released by hydrogen atom for transition $n=2$ to $n=1$

$\therefore \Delta E_{1}=13.6 \times\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\frac{3}{4} \times 13.6 \mathrm{eV}=10.2 \mathrm{eV}$

This energy is absorbed by $\mathrm{He}^{+}$ion in transition from $n$ $=2$ to $n=n_{1}$ (say)

$\therefore \Delta E_{2}=13.6 \times 4 \times\left(\frac{1}{4}-\frac{1}{n_{1}^{2}}\right)=10.2 \mathrm{eV}$

$\Rightarrow \mathrm{n}_{1}=4$

So, possible transition is $\mathrm{n}=2 \rightarrow \mathrm{n}=4$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now