What Excited State Was the Two-Times Ionized Lithium Initially In?

[skrat]

Homework Statement

Two times ionized Lithium (Z=3) in excited states emits two photons, one with $\lambda _1=72.91 nm$ and another one with $\lambda _2=13.5 nm$.
In which excited state was originally the ion?

Homework Equations

The Attempt at a Solution

Since two times ionized Lithium has only one electron I assume I can say that the energies of eigenstates are calculated as $E_n=\frac{R_y}{n^2}$.

I somehow imagined that I have to sum the energy of photons and equal that with the expression above.

$\frac{hc}{\lambda _1}+\frac{hc}{\lambda _2}+R_y=R_y(1-\frac{1}{n^2})$

But this gives me some weird result...

What do I have to do? :/

 

[DrClaude]

Since two times ionized Lithium has only one electron I assume I can say that the energies of eigenstates are calculated as $E_n=\frac{R_y}{n^2}$.

I somehow imagined that I have to sum the energy of photons and equal that with the expression above.

That's not correct. The energy of one photon already corresponds to a transition between two levels: the difference between the initial excited state and the final state.

Start again from the equation for the energy, but consider one transition at a time. After that you will have to combine both transitions to get the desired solution.

 

[skrat]

Than first transition is:

$\frac{hc}{\lambda _1}=R_y(\frac{1}{(n')^2}-\frac{1}{(n)^2})$

and second

$\frac{hc}{\lambda _2}=R_y(1-\frac{1}{(n')^2})$

I hope that's exactly what you had in mind.

However, to get rid of $(n')^2$ I woud simply sum the both equation.

$\frac{hc}{\lambda _2}+\frac{hc}{\lambda _1}=R_y(1-\frac{1}{n^2})$

 

[DrClaude]

$\frac{hc}{\lambda _2}+\frac{hc}{\lambda _1}=R_y(1-\frac{1}{n^2})$

That looks ok. What do you get when you try to solve this for $n$?

(Note: the problem is not clearly stated, and I misread it the first time. This explains why I might have made you rederive the equation you had at the beginning, although now without the extra $R_y$.)

 

[skrat]

The next part confuses me the most...

$n^2=\left [1-\frac{hc}{R_y}(\frac{1}{\lambda _2}+\frac{1}{\lambda _1}) \right ]^{-1}$

which gives me $n=0.378$

 

[DrClaude]

The next part confuses me the most...

$n^2=\left [1-\frac{hc}{R_y}(\frac{1}{\lambda _2}+\frac{1}{\lambda _1}) \right ]^{-1}$

which gives me $n=0.378$

You're using the wrong Rydberg constant. It has to be scaled for the charge of the nucleus.

 

[skrat]

You're using the wrong Rydberg constant. It has to be scaled for the charge of the nucleus.

$R_y=13.6 eV$

How do I do that?

Lithum has 3 nucleus is than the right Rydberg constant multiplied by 3?

 

[DrClaude]

$R_y=13.6 eV$

How do I do that?

Lithum has 3 nucleus is than the right Rydberg constant multiplied by 3?

Do you have the equation for the energy levels of hydrogenic atoms? See http://en.wikipedia.org/wiki/Hydrogen-like_atom

 

[skrat]

Ok, that's good to now.

So using $Z^2R_y$ gives me $n=3$.

Thank you!