Concept: In characteristic X-ray spectra, the wavelength of the emitted photon is inversely proportional to the energy difference between the transitioning electron shells.
Formula:$$ E = \frac{hc}{\lambda} \implies \lambda \propto \frac{1}{\Delta E} $$
Solution:- A longer wavelength requires a smaller energy transition (\( \Delta E \)).
- \( K \)-series X-rays result from transitions down to the innermost K shell (\( n=1 \)), yielding massive energy drops and very short wavelengths.
- The \( M_\alpha \) line results from a transition from the N shell (\( n=4 \)) to the M shell (\( n=3 \)).
- Because energy levels get much closer together further from the nucleus, the energy difference between N and M is much smaller than any transition ending at K.
- Smallest \( \Delta E \) results in the longest wavelength \( \lambda \).
Why other options are incorrect:Options A, B, and C all terminate at the deeply bound K shell, resulting in high-energy, extremely short-wavelength X-ray photons.
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