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NEET 2004
Exam
NEET
Year
2004
Subject
Physics
Chapter
Atomic Physics

NEET 2004 · Physics · Atomic Physics

Energy E of a hydrogen atom with principal quantum number n is...

Energy EE of a hydrogen atom with principal quantum number nn is given by E=−13.6n2eVE=\frac{-13.6}{n^2} \mathrm{eV}. The energy of a photon ejected when the electron jumps for n=3n=3 state n=2n=2 state of hydrogen is approximately

Options

  1. A1.5eV1.5 \mathrm{eV}
  2. B0.85eV0.85 \mathrm{eV}
  3. C3.4eV3.4 \mathrm{eV}
  4. D1.9eV1.9 \mathrm{eV}

Correct answer

D. 1.9eV1.9 \mathrm{eV}

Step-by-step solution

Energy of photon =E2−E2=E_2-E_2 \begin{gathered} =\frac{13.6}{9}-\left(\frac{-13.6}{9}\right)=\frac{5}{30} \times13.6 \\ =1.9 \mathrm{eV} \end{gathered}
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