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# An energy of about 21 eV is required to excite an electron in a helium atom from the 1s state to the 2s state. The same transition for the $\mathrm{He}^{+}$ ion requires about twice as much energy. Explain why this is so.

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As we know, the expression for the transition energy is given by:

\begin{align*} E_{n}&=\frac{2^2}{n^2} (-13.6 \: \text{eV}) \tag{Where is n the number of states.}\\ \end{align*}

The transition energy for $\text{He}^ {+}$ is about twice as He. In a neutral atom, one electron is moving in the field of a nucleus and the field of another electron. The nucleus contains two protons of charge $+ze$ it moves in the field of a net charge.

\begin{align*} q_{net}&=+2e-e\\ q_{net}&=+e\\ \end{align*}

The electron in the atom $\text{He}^{+}$ moves in the field of $ze$. So, the potential energy function for the electron is about double that of one electron in the neutral atom.

For He atom, the transition energy is:

$E_{2}-E_{1}=21 \: \text{eV}$

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