How Do You Calculate Ionization Energies for Different Shells in X-ray Spectra?

AI Thread Summary
The discussion focuses on calculating ionization energies for the L, M, and N shells in tungsten's x-ray spectrum. The K-shell ionization energy is given as 69.5 keV, and the participant successfully calculated the L and M shell energies as 11.7 keV and 10.0 keV, respectively. However, the calculation for the N shell resulted in an incorrect answer, initially calculated as 2.3 keV. After suggestions to use more significant figures in constants, the participant recalculated and found the correct answer to be 2.5 keV, highlighting the importance of precision in calculations involving similar magnitudes. The discussion emphasizes the need for accuracy in significant figures when performing such calculations.
NewSoul
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Homework Statement


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.

Homework Equations


E=\frac{hc}{\lambda}<br /> \\E_{n}=-\frac{(13.6\textrm{ eV})Z_\textrm{eff}^2}{n^2}<br /> \\1\textrm{ eV}=1.6\textrm{E-19 J}

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}\\<br /> 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}\\<br /> 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!
 
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NewSoul said:
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}
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.
 
haruspex said:
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.

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?
 
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
 
ehild said:
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

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.
 
NewSoul said:
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.
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.
 
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