The energy of He^+ ion in its first excited state is, (The ground… — Solution | JEE Physics

MCQEasyJEE 2023Bohr's Model & Hydrogen Spectrum

The energy of He $^+$ ion in its first excited state is, (The ground state energy for the Hydrogen atom is $-13.6 \, \text{eV}$ ):

Answer

Correct answer:A

Standard Method

Given: He $^+$ ion, first excited state so $n = 2$ , and the ground state energy for hydrogen is $-13.6 \, \text{eV}$ .

Find: The energy of He $^+$ ion in its first excited state.

For hydrogen-like ions, the energy levels are given by

E_n = -\frac{13.6 Z^2}{n^2} \, \text{eV}

Here, for He $^+$ , $Z = 2$ .

For the first excited state, $n = 2$ . Substituting these values,

E_2 = -\frac{13.6 \cdot (2)^2}{(2)^2}

So,

E_2 = -13.6 \, \text{eV}

Therefore, the energy of the He $^+$ ion in its first excited state is $-13.6 \, \text{eV}$ . The correct option is A.

Using the hydrogen value directly without applying the $Z^2$ factor is incorrect because He $^+$ is a hydrogen-like ion with nuclear charge $Z = 2$ . Always use $E_n = -\frac{13.6 Z^2}{n^2} \, \text{eV}$ .
Taking the first excited state as $n = 1$ is wrong because the ground state is $n = 1$ and the first excited state is $n = 2$ . Identify the quantum number before substitution.
Forgetting that the energy remains negative is a conceptual error because bound states of hydrogen-like ions have negative energy. Keep the negative sign throughout the calculation.

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