JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2019
- Medium
- MCQ
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
Pick an option to check your answer.
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 \(\mathrm{N} \rightarrow \mathrm{L}\)
\(\quad \frac{1}{\lambda^{\prime}}=\mathrm{K}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{\mathrm{K} \times 3}{16}\)
\(\lambda^{\prime}=\frac{20}{27} \lambda\)
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