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NEET PHYSICSATOMSEasy

Question

Energy E of a hydrogen atom with principal quantum number n is given by E = -13.6/n² eV. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen, is approximately:

A

1.5 eV

B

0.85 eV

C

3.4 eV

D

1.9 eV

Step-by-Step Solution

The energy of the emitted photon corresponds to the difference in energy between the two states: \Delta E = E₃ - E₂.

  1. Calculate energy of initial state (n=3): E₃ = -13.6 / 3² = -13.6 / 9 ≈ -1.51 eV.
  2. Calculate energy of final state (n=2): E₂ = -13.6 / 2² = -13.6 / 4 = -3.4 eV.
  3. Calculate the energy difference (energy of ejected photon): \Delta E = E₃ - E₂ = -1.51 - (-3.4) = 3.4 - 1.51 = 1.89 eV.

Rounding to one decimal place, the energy is approximately 1.9 eV.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ATOMS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSATOMSenergyhydrogenprincipalquantumnumber

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