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JEE Main 2025 (3 Apr, Shift 2) · Transitions, Photon Emission & Excitation

Transitions, Photon Emission & Excitation question from JEE Main 2025 (3 Apr, Shift 2)

Q16JEE Main 2025 · 3 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation

An electron in the hydrogen atom initially in the fourth excited state makes a transition to nnth energy state by emitting a photon of energy 2.86 eV. The integer value of n will be__________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Fourth excited state is n=5n = 5: E5=−13.625=−0.544E_5 = -\frac{13.6}{25} = -0.544 eV.
  2. En=E5−2.86=−3.40E_n = E_5 - 2.86 = -3.40 eV
  3. −13.6n2=−3.40⇒n2=4-\frac{13.6}{n^2} = -3.40 \Rightarrow n^2 = 4
  4. So n=2n = 2.
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