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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 zdirection
Relevant 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
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