Question 183: Atoms & Nuclei

Question

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

Accepted Answer

The energy levels of a one-electron ion can be described by the formula:
$$
E_n = -\frac{Z^2}{n^2} 	imes E_0
$$
where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom (-13.6 eV).
For the $$\mathrm{He}^{+}$$ ion, the atomic number Z is 2 (since helium has 2 protons). We are looking for the energy of the first excited state, which corresponds to n = 2. Plugging these values into the formula, we get:
$$
E_2 = -\frac{2^2}{2^2} 	imes (-13.6 ~\mathrm{eV}) = -13.6 ~\mathrm{eV}
$$
So, the energy of the $$\mathrm{He}^{+}$$ ion in its first excited state is $$-13.6 ~\mathrm{eV}$$.

Answer

$$-13.6 ~\mathrm{eV}$$

Answer

$$-27.2 ~\mathrm{eV}$$

Answer

$$-3.4 ~\mathrm{eV}$$

Answer

$$-54.4 ~\mathrm{eV}$$

Atoms & Nuclei

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

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