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Electron excited from the ground state to a quantum state

  • Thread starter jisbon
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  • #1
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Homework Statement:

The electron in a He+ atom is excited from ground to quantum state n=3 by radiation.
i.Calculate the wavelength of the radiation
ii.Calculate the magnitude of orbital angular momentum
iii.Calculate the magnitudes of the possible orbital angular momentum in z-direction

Homework Equations:

##E_{n}=\dfrac {-13.6z^{2}}{n^{2}}eV##
Hi there, popping by here to check my answer because another online platform has already answered it but my answer appears to be wrong. I can't seem to understand why though :/

Since I can find the energy at a state to be ##E_{n}=\dfrac {-13.6z^{2}}{n^{2}}eV##
At ground state where n=1,
##E_{1}=\dfrac {-13.6(2)^{2}}{1^{2}}eV = -54.4eV##
And at n=3,
##E_{3}=\dfrac {-13.6(2)^{2}}{3^{2}}eV = -6.04444...eV##
Since photon energy ##E_{p}h=hf=\dfrac {hc}{\lambda }=\dfrac {1240}{\lambda }eV ##
Shouldn't it just be:
## -6.04444...eV-(-54.4eV) = \dfrac {1240}{\lambda }eV##
and the wavelength I got is 25.6nm, which is a far cry from the online solution which is 102nm :/

As for part ii, since ##L=\sqrt {l\left( l+1\right) }\hbar##, and l must be =1 due to the selection rule of allowed transitions, hence ##L=\sqrt {2}\hbar ##?

And lastly for part iii, since ##m_{l}## is -1,0,1 , ##L_{z}## must be ##\hbar## or -##\hbar##?

Cheers
 

Answers and Replies

  • #2
gneill
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and the wavelength I got is 25.6nm, which is a far cry from the online solution which is 102nm :/
Looks like they did the calculation for hydrogen, not helium.
 
  • #3
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Looks like they did the calculation for hydrogen, not helium.
So my answer/thought process is correct in this case? Thanks
 
  • #4
gneill
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So my answer/thought process is correct in this case? Thanks
Yup. Certainly looks that way.
 
  • #5
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Yup. Certainly looks that way.
Guess I can't trust all online solutions :/ Thanks for the clarification :D
 
  • #6
gneill
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You're most welcome. Here we regularly spot errors in other site's purported solutions.
 

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