JEE Mains · Physics · STD 12 - 12. atoms
In a hydrogen like atom, when an electron jumps from the \(M -\) shell to the \(L-\) shell the wavelength of emitted radiation is \(\lambda \). If an electron jumps from \(N-\) shell to the \(L-\) shell the wavelength of emitted radiation will be
- A \(\frac{{27}}{{20}}\,\lambda \)
- B \(\frac{{16}}{{25}}\,\lambda \)
- C \(\frac{{25}}{{16}}\,\lambda \)
- D \(\frac{{20}}{{27}}\,\lambda \)
Answer & Solution
Correct Answer
(D) \(\frac{{20}}{{27}}\,\lambda \)
Step-by-step Solution
Detailed explanation
For \(M \rightarrow L\) steel \(\frac{1}{\lambda}=\mathrm{K}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{K} \times 5}{36}\) for \(\mathrm{N} \rightarrow \mathrm{L}\)…
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