Concept: Photon wavelength is inversely proportional to the energy of the atomic transition. Longest wavelength equates to the lowest possible energy jump between shells.
Formula:$$ \Delta E = E_{\text{upper}} - E_{\text{lower}} = \frac{hc}{\lambda} $$
Solution:- The inner shells (like K, \( n=1 \)) have massive binding energies. Transitions ending at the K-shell release large amounts of energy (short wavelength).
- The M-shell (\( n=3 \)) is much further from the nucleus. An \( M_\alpha \) photon is generated by an electron falling from the N-shell (\( n=4 \)) to the M-shell.
- Because higher orbital energy levels are spaced very closely together, the energy difference between N and M is very small.
- This incredibly small energy gap generates the lowest energy photon among the options, and therefore the longest wavelength.
Why other options are incorrect:Options A, B, and C all represent electrons plunging deep into the intensely bound K-shell, releasing highly energetic, short-wavelength X-rays.
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