Visible Light!
A sample of H-atoms all in a lower energy level ( n 1 ) absorb monochromatic light of energy H e V to get excited to a Higher energy ( n 2 ) , subsecuently the atom emits radiation of 6 different photon energies in returning to ground state
Some of these emmited photons have Energy < x e V , some = x e V , and some > x e V .
When an electron in the same H-atom transits from energy level n 2 to n 1 , the colour of the light released is
1 . R E D
2 . V I O L E T
3 . B L U E − G R E E N
4 . I N D I G O
Enter your answer as sum of respective option numbers
e g : if your answer is 1,2,3,4 then enter 1+2+3+4 = 10.
TAKE UP THE CHALLENGES!!
This is a part of my set Aniket's Chemistry Challenges (BEST OF JEE - ADVANCED)
The answer is 4.
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1 solution
How do we know that 3-2 and 4-2 will give the respective colours?And if you have some list related to it please do share!
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| TRANSITION | COLOUR |
| 9 → 2 | Ultra Violet |
| 8 → 2 | Ultra Violet |
| 7 → 2 | Ultra Violet |
| 6 → 2 | Violet |
| 5 → 2 | Violet |
| 4 → 2 | Blue green (cyan) |
| 3 → 2 | Red |
| 3 → 2 | Red |
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Right! One change - 5 to 2 gives indigo ( to be more appropriate ) :)
Also, from 7 to 2 and 8 to 2 and 9 to 2 , I have dbt as I learnt Balmer ones are in visible region
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@Aniket Sanghi – Balmer α, Balmer β, Balmer γ and Balmer δ are the visible ones in the visible range, the rest are in UV range, so when we describe this series in brief along with the others, we say in general that it has wavelengths in the visible range. Check this... Balmer series- Wikipedia
I am facing a problem, if 6 different energies are emitted, there should be 6 different spectral lines, and for that the formula 2 ( n 2 − n 1 + 1 ) ( n 2 − n 1 ) gives n 2 = 5 . Where am I wrong?
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Put n1 = 1 , I think u are putting it something else
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Okay, now I got it, I was thinking that the question asked all possible wavelengths emitted, but it asks only the ones visible to human eye, so there would be 2 of them as mentioned 4 → 2 and 3 → 2. Your first statement in solution is true, the whole question is a confusion! :)
Full que is just a confusion....... main things to focus are....
emmission of 6 diff energies ...this imlies that n2 = 4.......
now since visible light is emmited when it transits fron n2 to n1 ....this implies that n1 = 2.....
possible transitions is 3-2 and 4-2.....hence answer is blue-green and red...