Bohr model question - transition between first excited state and ground state

  • Thread starter daleklama
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Homework Statement



In a hydrogenlike ion with atomic number Z, the energies of the allowed states are given by

E(n) = (-13.6eV) (Z^2/n^2)

What is the wavelength asociated with the transition between first excited state and ground state of hydrogen-like helium? (He+)

Homework Equations



E(n) = (-13.6eV) (Z^2/n^2)
E = hf = hc/Lambda

The Attempt at a Solution



Hydrogen like helium is He+ which has 2 protons and 1 electron. Atomic number Z is 2.
So I'd sub in 2 for Z (4 for Z^2), but I don't know what to choose n as.
I assume n=1 for ground state and n=2 for first excited state.
An idea I had was to do it twice - sub in n=1 and complete the equation, sub in n=2 and complete the equation, and... take them away?
I'm not sure :(

Thanks very much.
 

Answers and Replies

  • #2
881
40
Hydrogen like helium is He+ which has 2 protons and 1 electron. Atomic number Z is 2.
So I'd sub in 2 for Z (4 for Z^2), but I don't know what to choose n as.
I assume n=1 for ground state and n=2 for first excited state.
An idea I had was to do it twice - sub in n=1 and complete the equation, sub in n=2 and complete the equation, and... take them away?
I'm not sure :(

Thanks very much.
n is the principle quantum number, which goes like, n = 1 for ground state, 2 for 1st excited state, 3 for 2nd excited state and so on, so your assumption is correct. Now you'll have the energies for both the ground and first excited states. So, doing the second step you suggested would be correct, too. You need the difference in their energy states, after all.
 
  • #3
898
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Yes, calculate the difference between E(n=1) and E(n=2). That is the energy of the photon. Then find the wavelength that corresponds to that energy.
 
  • #4
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Thanks both, I understand :)

One question though, does it give me the (energy) answer in ev? I assume it does since ev is the unit of the -13.6.
And then, do I need to convert both to Joules, then take them away, and then find corresponding wavelength?

Thanks again.
 
  • #5
881
40
E(n) = (-13.6eV) (Z^2/n^2)
Z and n are just plain numbers, without any units, and 13.6 is in eV, so yep. :wink:


Thanks both, I understand :)

One question though, does it give me the (energy) answer in ev? I assume it does since ev is the unit of the -13.6.
And then, do I need to convert both to Joules, then take them away, and then find corresponding wavelength?

Thanks again.
This depends on which kind of values you are using for the equation [itex]E = hc/\lambda[/itex], i.e the unit of c and h. The answer units for [itex]\lambda[/itex] will vary accordingly. Just make sure all the variables, E, h, c have the same type of units, or you'll have to convert them. Its usually a good idea to stick to SI units, though.
 

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