Q134 / 200 · NEET 2010 (Screening) · Physics

The energy of a hydrogen atom in the ground state is $-13.6 \mathrm{eV}$. The energy of a $\mathrm{He}^{+}$ion in the first excited state will be

  • Correct answer
    A
    $-13.6 \mathrm{eV}$
  • B
    $-27.2 \mathrm{eV}$
  • C
    $-54.4 \mathrm{eV}$
  • D
    $-6.8 \mathrm{eV}$

Solution

Energy E of an atom with principal quantum number $\mathrm{n}$ is given by $E=\frac{-13.6}{\mathrm{n}^2} Z^2$ for first excited state $\mathrm{n}=2$ and for $\mathrm{He}^{+} \mathrm{Z}=2$ $$ \begin{aligned} \Rightarrow \quad \mathrm{E} & =\frac{-136 \times(2)^2}{(2)^2} \\ & =-13.6 \mathrm{eV} \end{aligned} $$

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

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