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Calculate the approximate energy of an X-ray emitted for an n = 2 to n = 1 transition in a tungsten anode, where the effective atomic number is 73. Use En = -(Z^2 / n^2)(13.6 eV).
A18.1 keV
B36.2 keV
C54.4 keV
D72.5 keV
Answer & Solution
Correct answer: C. 54.4 keV
1. The n = 1 energy is Ef = -(73 squared divided by 1 squared)(13.6 eV), which is about -72.5 keV.
2. The n = 2 energy is Ei = -(73 squared divided by 2 squared)(13.6 eV), which is about -18.1 keV.
3. The emitted energy is the initial energy minus the final energy: -18.1 minus (-72.5) keV.
4. Subtracting gives 54.4 keV.
5. 18.1 keV and 72.5 keV are the individual level energies, not their difference, and 36.2 keV does not match this subtraction.
6. So 54.4 keV is correct.
_Source: OpenStax Physics (CC BY 4.0), Ch 22 "The Atom", section 22.1 The Structure of the Atom_
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