Electromagnetic emission lines for a hydrogen atom

  • #1
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Homework Statement


Hi, I've been unable to find a relevant thread for a question that I've been stuck on for a couple of days now.

Here it is;

One of the electromagnetic emission lines for a hydrogen atom has wavelength 389nm. Assiming that this is a line from one of the Lyman (nf =1 ), Balmer (nf = 2) or Paschen (nf =3) series, what is the initial principle quantum number associated with the transitions? The Rydberg energy can be assumed to be 13.6eV



Homework Equations


ΔE=Ry(1/(ni)2 - 1/(nf)2))
Ry=13.6eV

This is the formula used to attempt a solution

There is another formula in my notes that may be helpful but i do not understand the symbols within it, it is;
En,l = -Ry/(n-δl)2

The Attempt at a Solution



Using the equation ΔE=Ry(1/(ni)2 - 1/(nf)2)) and then equating this change in energy to the energy of light E=hc/λ.
Ry=13.6eV
Then I insert various nf 's e.g. nf =1 for Lyman, then i would solve this for ni to see if i get an integer number, indicating that this is correct. I do this for nf =1,2 and 3 and I have never got an integer number leading me to believe my method is incorrect.

Any help or insight would be appreciated,

Thanks in advance, John.
 

Answers and Replies

  • #3
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I forgot to mention doing these questions are purely for exam practice. So having the tables will no be an option.

How would i do this in an exam? Simply put the numbers in as i have done and then chose the one which is closest to an integer value to be the correct one?

Thank, John
 
  • #4
TSny
Homework Helper
Gold Member
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Using the equation ΔE=Ry(1/(ni)2 - 1/(nf)2))
Check your equation in regard to the placement of ni and nf.

I think your approach to the problem is good.
 
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Likes Jdraper
  • #5
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Ahh, yes, the ni and nf are the wrong way around. Thanks for your help :)
 

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