Q95 / 160 · JIPMER 2013 · Physics

Minimum excitation potential of Bohr's first orbit in hydrogen atom is

  • A
    3.6 V
  • Correct answer
    B
    10.2 V
  • C
    13.6 V
  • D
    3.4 V

Solution

From the relation $\begin{aligned} \Delta E & =13.6 \mathrm{eV}-\frac{13.6 \mathrm{eV}}{n^2} \\ & =13.6 \mathrm{eV}-\frac{13.6 \mathrm{eV}}{(2)^2}=10.2 \mathrm{eV}\end{aligned}$ Therefore, excitation potential $=\frac{10.2}{e} \mathrm{eV}=10.2 \mathrm{~V}$

Hence, the correct answer is option: 'B'.

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