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JEE Main 2002
Exam
JEE Main
Year
2002
Subject
Physics
Chapter
Atomic Physics

JEE Main 2002 · Physics · Atomic Physics

If 13.6 eV energy is required to ionize the hydrogen atom, then the...

If 13.6eV13.6 \mathrm{eV} energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n=2n=2 is

Options

  1. A10.2eV10.2 \mathrm{eV}
  2. B0eV0 \mathrm{eV}
  3. C3.4eV3.4 \mathrm{eV}
  4. D6.8eV6.8 \mathrm{eV}

Correct answer

C. 3.4eV3.4 \mathrm{eV}

Step-by-step solution

En=−13.6n2⇒E2=−13.622=3.4eVE_n=-\frac{13.6}{n^2} \Rightarrow E_2=-\frac{13.6}{2^2}=3.4 \mathrm{eV}
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