Calculation of no. of spectral lines for group of similar atoms

[Prabhu1]

Homework Statement

The maximum no of spectral lines for a single atom during it's electron's transition is given by [∆n(∆n+1)]/2 . But I don't seem to arrive at the answer when a group of atoms are present . The question was - What is the maximum number of spectral lines possible for Balmer series when a set of 8 hydrogen atoms are irradiated with light and all are excited to 6th excited state and spectrum is obtained?
2. Homework Equations
[∆n(∆n+1)]/2

The Attempt at a Solution

.[/B]
I am able to calculate the spectral lines for 1 atom which comes out to be 15 . But the answer key says 5 . i am not able to understand how .

 

[mfb]

Is that the exact full problem statement? If yes, I don't understand the given answer.

 

[DrClaude]

Your equation tells you about the total possible spectral lines, but the problem asks about the Balmer series.

Also, be careful about the wording of the problem.. What is the 6th excited state?

 

[Prabhu1]

Your equation tells you about the total possible spectral lines, but the problem asks about the Balmer series.

Also, be careful about the wording of the problem.. What is the 6th excited state?

The 7th state

 

[DrClaude]

The 7th state

Correct. So how many Balmer series lines do you get?

 

[Prabhu1]

Correct. So how many Balmer series lines do you get?

If we come down from each of the higher states directly like 7 ->2 , 6->2 and so on, we will get 5 lines , but there are still other ways to reach the 2nd state , aren't there?

 

[DrClaude]

but there are still other ways to reach the 2nd state , aren't there?

Like what?

 

[Prabhu1]

Like what?

7 -> 6 ->2 .

 

[DrClaude]

7 -> 6 ->2 .

Wasn't that already taken care of who you wrote

If we come down from each of the higher states directly like 7 ->2 , 6->2 and so on, we will get 5 lines

In other words, where would the 6->2 come from if not from 7->6?

 

[Prabhu1]

Wasn't that already taken care of who you wroteIn other words, where would the 6->2 come from if not from 7->6?

But still , we are getting a new spectral line for transition from 7 ->6 .

 

[Prabhu1]

But still , we are getting a new spectral line for transition from 7 ->6 .

But that wouldn't be included in balmer series , right?

 

[DrClaude]

But still , we are getting a new spectral line for transition from 7 ->6 .

Is that line part of the Balmer series?

 

[DrClaude]

But that wouldn't be included in balmer series , right?

Our messages crossed. Yes, that's the point. You should only count the lines that end on n=2.

 

[Prabhu1]

Our messages crossed. Yes, that's the point. You should only count the lines that end on n=2.

Okay , got you , silly mistake I did there .thank a ton sir . Btw I really liked your way of helping me , you didn't give out the answer straight away . thanks again .

 

[DrClaude]

Okay , got you , silly mistake I did there .thank a ton sir . Btw I really liked your way of helping me , you didn't give out the answer straight away . thanks again .

This is the way we do things here.

 

[gaurav singh]

But aren't there 8 hydrogen atoms, that will make it 8X5=40

 

[mfb]

The atoms are all identical, they have identical spectral lines.