# Finding ionization energy

1. Apr 11, 2014

### NewSoul

1. The problem statement, all variables and given/known data
The K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV.
(a) Determine the ionization energies of the L, M, and N shells.

2. Relevant equations
$$E=\frac{hc}{\lambda} \\E_{n}=-\frac{(13.6\textrm{ eV})Z_\textrm{eff}^2}{n^2} \\1\textrm{ eV}=1.6\textrm{E-19 J}$$

3. The attempt at a solution
I correctly determined the ionization energies for the L and M subshells (11.7 and 10.0 keV respectively. However, I am calculating the value for the N subshell in the exact same way, but my homework tells me the answer is incorrect.

Here are my calculations for the L shell...
$\lambda$ of L =.0215E-9 m
$$E_\textrm{KL}=\frac{(6.63\textrm{E-34})(3\textrm{E8})}{.0215\textrm{E-9}}=9.25\textrm{E-15 J}=57800\textrm{ eV}=57.8\textrm{ keV}\\ E_\textrm{L}=69.5-57.8=11.7\textrm{ keV}$$
I did the same calculations to find the ionization energy for subshell M of 10.0 keV. Both of these answers were correct.

For N, I'm doing the exact same thing again and winding up with what my online homework says to be the wrong answer. I do not know the correct answer.

My work for N...
$$E_\textrm{KN}=\frac{(6.63\textrm{E-34})(3\textrm{E8})}{.0185\textrm{E-9}}=1.08\textrm{E-14 J}=672800\textrm{ eV}=67.2\textrm{ keV}\\ E_\textrm{N}=69.5-67.2=2.3\textrm{ keV}$$
It says my answer is close, but incorrect. I also tried 2.4 in case of a round off error, but that was also wrong. I'm not sure what's up.

Thanks so much!

Last edited: Apr 12, 2014
2. Apr 11, 2014

### haruspex

Wouldn't 67.3 be closer? That would give 2.2 as the answer.
Btw, is your notation right? Compared with what I would have expected, you seem to have interchanged EKL with EL and EKN with EN.

3. Apr 12, 2014

### NewSoul

Whoops, I do have those two switched!

2.2 didn't work either. I honestly have no idea what the problem is. Nothing obvious, eh?

4. Apr 12, 2014

### ehild

Use more significant digits in the constants h, c, e: c=2.998 E8, h=6.628 E-34, 1eV=1.602 E-19 J. Do not round the results for the photon energies. Round off at the end.

ehild

5. Apr 12, 2014

### NewSoul

Thanks. To be safe, I went to 9 significant figures and ended up with the apparently correct answer of 2.5 keV. I just don't understand why it wouldn't accept my other answer. We've been using this online program for a while now and it has always been fine when I don't use that many significant figures for the constants.

6. Apr 12, 2014

### haruspex

The problem here is that in the working you had to take the difference of two numbers of similar magnitude. That resulted in one less (in fact, 1.5 less) significant figure of precision than you started with.