How Many Different Photon Energies Are Emitted from Hydrogen Atoms in n=5 State?

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SUMMARY

The discussion focuses on the emission of photon energies from hydrogen atoms in the n=5 state transitioning to the ground state (n=1). It is established that there are 10 possible downward transitions, calculated using the combination formula C(5,2), which accounts for all unique pairs of energy levels. The confusion arises regarding the relationship between the number of hydrogen atoms and the total number of emitted photons, with clarification that the expected number of photons emitted for one atom can be multiplied by the total number of atoms (500) to find the overall emission.

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Homework Statement



A sample of hydrogen atoms are all in the n=5 state. I all the atoms return to the ground state, how many different photon energies will be emitted , assuming all possible transitions occur? If there are 500 atoms in the sample and assuming that from any state all possible downward transitions are equally probably, what is the total number of photons that will be emitted when all atoms return to the ground state

Homework Equations


I don't think there is an equation.

The Attempt at a Solution



In the first part of the problem, wouldn't the ground state, be n=1. If true, then wouldn't there be four different photon energies emitted? Energies would be emitted from n=5 to n=1, Other energies would be emitted from n=5 to n=2 , n=5 to n=3 , and n=5 to n=4. I'm not sure how to figure out the second answer to the problem.

Actually for the first part of the problem, the back of my book says that there are 10 possible transitions instead of 4 . I do not understand why there are 10 possible transitions
 
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please help me.
 
Benzoate said:
please help me.

The answer is 5 choose 2... which is 5!/(2!3!)

That's considering all the transitions to a lower level:

ie:

5 ->4
5 ->3
5 ->2
5 ->1

4 ->3
4 ->2
4 ->1

3 ->2
3 ->1

2 ->1

But like yourself, I'm confused why they say "returns to the ground state" and then include transitions like 4->3 which don't go to the ground state...

EDIT: I think they mean "eventually" return to the ground state... ie not return to the ground state in a single transition.
 
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The "returns to the ground state" is for the second part of the problem... not the first.
 
learningphysics said:
The "returns to the ground state" is for the second part of the problem... not the first.

I don't want you to solved the second part of the problem for me, I'm just having trouble relating the total number of hydrogen atoms in the sample to the total number of transitions that take place.
 
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Benzoate said:
I don't want you to solved the second part of the problem for me, I'm just having trouble relating the total number of hydrogen atoms in the sample to the total number of transitions that take place.

It is not related to the number of hydrogen atoms. That's not needed for the first part of the problem.

10 is just the total number of possible downward transitions two states chosen from {1,2,3,4,5}

It is just 5 choose 2... ie C(5,2)... combinations...

I've listed all the transitions in my previous post...
 
learningphysics said:
It is not related to the number of hydrogen atoms. That's not needed for the first part of the problem.

10 is just the total number of possible downward transitions two states chosen from {1,2,3,4,5}

It is just 5 choose 2... ie C(5,2)... combinations...

I've listed all the transitions in my previous post...

I think there been a misunderstanding. I understand how you calculated the total number of transitions. I'll post the second part of the problem:

If there are 500 atoms in the sample and assuming that from any state all possible downward transitions are equally probably, what is the total number of photons that will be emitted when all atoms return to the ground state

I don't understand how to related the 500 atoms in the sample to the total number of the transitions.
 
The idea is to get the "expected number of photons emitted" for 1 atom...

Then the number of photons emitted for 500 atoms, is just 500 times that...

The main part of the problem is to get the average number of photons emitted for 1 atom.
 

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