Question

In a mixture of sample of H-atoms and He+ ions, electrons in all the H-atoms and He+ ions are present in 4th state. Then, find maximum number of different spectral lines obtained when all the electrons make transitions from 4th state upto ground state

Answer 1

The total number of spectral lines = 1 + 6 + 15 = 22 lines.

Explanation:

The hydrogen atoms are in the first excited state. So n = 2. The only possible transition is n = 2 → n = 1.

n = 1 is the ground state and is the lowest energy level. This gives one spectral line.  

The helium ions are in the 3rd excited stated. So n = 4. The possible transitions are:  

4→3  

4→2  

4→1  

3→2  

3→1  

2→1  

We find that there are 6 different transitions which will show 6 different lines in the spectra. There are 6 spectral lines.

The lithium ions are in the 5th excited stated. So n = 6. The possible transitions are:  

6→5  

6→4  

6→3  

6→2  

6→1  

5→4  

5→3  

5→2  

5→1  

4→3  

4→2  

4→1  

3→2  

3→1  

2→1  

We find that there are 15 different transitions which will show 6 different lines in the spectra. There are 15 spectral lines.

The energy levels in lithium ions, hydrogen ions and helium ions are different from each other, so the total number of spectral lines = 1 + 6 + 15 = 22 lines.