Q137 / 180 · KCET 2021 · Physics

Energy of an electron in the second orbit of hydrogen atom is $E_{2}$. The energy of electron in the third orbit of $\mathrm{He}^{+}$will be

  • A
    $\frac{9}{16} E_{2}$
  • Correct answer
    B
    $\frac{16}{9} E_{2}$
  • C
    $\frac{3}{16} E_{2}$
  • D
    $\frac{16}{3} E_{2}$

Solution

The energy of the electron in the $n$th orbit is given by $E_{n}=-13.6 \mathrm{eV} \frac{Z^{2}}{n^{2}}$ $E_{n} \propto \frac{Z^{2}}{n^{2}}$ For hydrogen atom, $Z=1$ $E_{n} \propto \frac{1}{n^{2}}$ In second orbit, $n=2$ $E_{2} \propto \frac{1}{(2)^{2}}$ In case of $\mathrm{He}^{+}, Z=2$ and $n=3$ $\begin{aligned} & E_{3} \propto \frac{(2)^{2}}{(3)^{2}} \\ \therefore \quad & \frac{E_{2}}{E_{3}}=\frac{1}{4} \times \frac{9}{4}=\frac{9}{16} \\ \Rightarrow \quad & E_{3}=\frac{16 E_{2}}{9} \end{aligned}$

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

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