# Energy of photons from Balmer series

1. Mar 7, 2013

### Fluxthroughme

I can see how I would do part b, but I am confused as to why I have got part a wrong. Apparently the answer is 12.09eV, but I cannot see how. It seems to me that using the formula as I did should give the answer. Thanks for any help.

2. Mar 7, 2013

### ap123

You've worked out the energy of the photon emitted going from n=3 to n=2.
It wants the energy required to go from the ground level to to n=3 ( ie to the level where it can emit the Hα photon.

3. Mar 7, 2013

### Fluxthroughme

But if I do $\frac{1}{\lambda_\alpha} = R(\frac{1}{1^2} - \frac{1}{3^2})$, I'm not using the Balmer series, am I? I thought it needed to be a $2^2$ for the Balmer series? If I use a 1, how can it be the $H_\alpha$ line?

Unless you're saying the $H_\alpha$ is emitted during the transition from n=3 to n=2? Thus, it wants the energy to get to the n=3 in the first place, rather than the energy of the photon that would give the $H_\alpha$?

4. Mar 7, 2013

### ap123

Yes (this agrees with the supplied answer.)

5. Mar 7, 2013

### Fluxthroughme

Ahh, ok. This sorts out my confusion, then. Thank you for the help.