Question:

The ground state energy of hydrogen atom is \(-13.6\) eV. The energy of second excited state of \(\mathrm{He^{+}}\) ion in eV is :

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Use \(E_n = -13.6\,Z^2/n^2\) eV with Z = 2 and the second excited state n = 3.
Updated On: Nov 5, 2026
  • \(-6.04\)
  • \(-54.4\)
  • \(-27.2\)
  • \(-3.4\)
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The Correct Option is A

Solution and Explanation

For a hydrogen-like ion, \(E_n = -13.6\,\frac{Z^2}{n^2}\) eV.
For \(\mathrm{He^{+}}\), Z = 2. Ground state n = 1, first excited n = 2, second excited n = 3.
\(E_3 = -13.6 \times \frac{4}{9} = -6.04\) eV.
Answer: (1) \(-6.04\)
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