# Balmer Series for ##Li^{2+}##

## Homework Statement

The emission wavelengths of hydrogen-like atoms are related to nuclear charge. How do they scale as a function of Z? What are the longest and shortest wavelengths in the Balmer series for ##Li^{2+}##?

## Homework Equations

##E_n = -\frac{R}{n^2}## (1)

##a_0 = \frac{\hbar^2}{Zme^2}## (2)

From (2), ##\hbar^2 = a_0Zme^2## (3)

##\alpha = \frac{Ze^2}{\hbar c}## (4)

##R = \frac{1}{2} mc^2 \alpha^2## (5)

## The Attempt at a Solution

I started by trying to find an expression for the energy levels of hydrogen-like atoms. Substituting (5) into (1) gives
##E_n = -\frac{mc^2\alpha^2}{2n^2}##
Substitute in (4):
##E_n = -\frac{mZ^2e^4}{2\hbar^2 n^2}##
Sub in (3):
##E_n = -\frac{mZ^2e^4}{2Za_0me^2 n^2}##

Simplifying, ##E_n = -\frac{Ze^2}{2a_0n^2}##. This can be related to wavlength via ##E_n = \frac{hc}{\lambda}##, so emission and absorption wavelengths of hydrogenic atoms are related to nuclear charge Z by the function

##\lambda = -\frac{2hca_0n^2}{Ze^2}##

But I have no idea how to answer the bit about the Balmer series because n=2 and Z=3, so based on my equation isn't there only going to be one answer? How do I get minimum/maximum wavelengths?

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But I have no idea how to answer the bit about the Balmer series because n=2 and Z=3, so based on my equation isn't there only going to be one answer? How do I get minimum/maximum wavelengths?
In the Balmer series the final state is an ##n=2## state. The initial state may be anything from ##n=3## to ##n=\infty##. You need to find the energy difference between these states.

• Kara386
In the Balmer series the final state is an ##n=2## state. The initial state may be anything from ##n=3## to ##n=\infty##. You need to find the energy difference between these states.
So I use the equation for ##E_n## I found and do ##E_3-E_2## for the maximum wavelength? This gives a negative answer though, can I just multiply that answer by ##-1##? Or do ##E_2-E_3##?

##E_3-E_2## is positive from your formula since you got a minus sign in it. That is ##-1/9 > -1/4##.
I don't know if all your other constants are correct but you can always compare your answer to the Rydberg formula https://en.wikipedia.org/wiki/Rydberg_formula

• Kara386