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Physics · Atomic Physics

JEE Main 2025 — 24 January, Morning Shift — Question 51

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000A2000 A and it becomes 6000A6000 A when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-

Question figure
  1. Option A:

    3000A3000 A

    Correct
  2. Option B:

    6000A6000 A

  3. Option C:

    4000A4000 A

  4. Option D:

    2000A2000 A

Answer: A

Step-by-step solution

EA−EC=hc2000A….\mathrm{E}_{\mathrm{A}}-\mathrm{E}_{\mathrm{C}}=\frac{\mathrm{hc}}{2000 A} \ldots .. (i) and EB−EC=hc6000AE_{B}-E_{C}=\frac{h c}{6000 A} Now

EA−EB=(EA−EC)−(EB−EC)E_{A}-E_{B}=\left(E_{A}-E_{C}\right)-\left(E_{B}-E_{C}\right) hcλAB=hc2000−hc6000\frac{\mathrm{hc}}{\lambda_{\mathrm{AB}}}=\frac{\mathrm{hc}}{2000}-\frac{\mathrm{hc}}{6000} 1λAB=13000A\frac{1}{\lambda_{A B}}=\frac{1}{3000 A} λAB=3000A\lambda_{A B}=3000 A

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom

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