Electron excited from the ground state to a quantum state

[jisbon]

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

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

 

[gneill]

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.

 

[jisbon]

Looks like they did the calculation for hydrogen, not helium.

So my answer/thought process is correct in this case? Thanks

 

[gneill]

So my answer/thought process is correct in this case? Thanks

Yup. Certainly looks that way.

 

[jisbon]

Yup. Certainly looks that way.

Guess I can't trust all online solutions :/ Thanks for the clarification :D

 

[gneill]

You're most welcome. Here we regularly spot errors in other site's purported solutions.