# Finding ionization energy

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
$$E_\textrm{L}=\frac{(6.63\textrm{E-34})(3\textrm{E8})}{.0215\textrm{E-9}}=9.25\textrm{E-15 J}=57800\textrm{ eV}=57.8\textrm{ keV}\\ E_\textrm{KL}=69.5-57.8=11.7\textrm{ keV}$$
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…
$$E_\textrm{N}=\frac{(6.63\textrm{E-34})(3\textrm{E8})}{.0185\textrm{E-9}}=1.08\textrm{E-14 J}=672800\textrm{ eV}=67.2\textrm{ keV}\\ E_\textrm{KN}=69.5-67.2=2.3\textrm{ keV}$$
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!

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