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
[tex]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}[/tex]

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…
[itex]\lambda[/itex] of L =.0215E-9 m
[tex]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}[/tex]
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…
[tex]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}[/tex]
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!

http://ift.tt/1lWgiH5

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons