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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2019 · 11 January · Evening Shift · Physics · Atomic Physics

In a hydrogen like atom, when an electron jumps from the M-shell to...

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is LL. If an electron jumps from N\mathrm{N} -shell to the L\mathrm{L} -shell, the wavelength of emitted radiation will be:

Options

  1. A2720λ\frac{27}{20} \lambda
  2. B1625λ\frac{16}{25} \lambda
  3. C2516λ\frac{25}{16} \lambda
  4. D2027λ\frac{20}{27} \lambda

Correct answer

D. 2027λ\frac{20}{27} \lambda

Step-by-step solution

When electron jumps from M→L\mathrm{M} \rightarrow \mathrm{L} shell 1λ=K(122−132)=K×536\frac{1}{\lambda}=\mathrm{K}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{K} \times 5}{36} When eletron jumps from N→L\mathrm{N} \rightarrow \mathrm{L} shell 1λ′=K(122−142)=K×316\frac{1}{\lambda^{\prime}}=K\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{K \times 3}{16} solving equation (i) and (ii) we get λ′=2027λ\lambda^{\prime}=\frac{20}{27} \lambda
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