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A singly ionized helium atom in an excited state $(n=4)$ emits a photon of energy $2.6 \mathrm{eV}$. Given that the ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$, the energy $\left(\mathrm{E}_{\mathrm{t}}\right)$ and quantum number (n) of the resulting state are respectively,
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The correct answer is:
$E_{t}=-6.0 \mathrm{eV}, \mathrm{n}=3$
$$
\begin{array}{l}
2.6=13.6 z^{2}\left[\frac{1}{n^{2}}-\frac{1}{4^{2}}\right] \\
\frac{2.6}{13.6 \times 4}=2^{2}\left[\frac{1}{n^{2}}-\frac{1}{4^{2}}\right]
\end{array}
$$
$$
\frac{2.6}{13.6 \times 4}=\frac{1}{n^{2}}-\frac{1}{16}
$$
$n=3$ now Energy
$$
\begin{array}{l}
E=\frac{13.6 Z^{2}}{n^{2}} e V \\
-\frac{13.6 \times 4}{9} e V=-6 e V
\end{array}
$$
\begin{array}{l}
2.6=13.6 z^{2}\left[\frac{1}{n^{2}}-\frac{1}{4^{2}}\right] \\
\frac{2.6}{13.6 \times 4}=2^{2}\left[\frac{1}{n^{2}}-\frac{1}{4^{2}}\right]
\end{array}
$$
$$
\frac{2.6}{13.6 \times 4}=\frac{1}{n^{2}}-\frac{1}{16}
$$
$n=3$ now Energy
$$
\begin{array}{l}
E=\frac{13.6 Z^{2}}{n^{2}} e V \\
-\frac{13.6 \times 4}{9} e V=-6 e V
\end{array}
$$
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