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What is the degeneracy of the level of H-atom that has energy $\left(-\frac{R_{H}}{9}\right)$ ?
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The correct answer is:
9
Energy of the electron in the nth orbit in terms of $\mathrm{R}_{\mathrm{H}}$ is
$\mathrm{E}_{\mathrm{n}}=-\frac{\mathrm{R}_{\mathrm{H}} \mathrm{Z}^{2}}{\mathrm{n}^{2}}$
where, $Z=$ atomic number, $n^{2}=$ degeneracy
For $\mathrm{H}$-atom, $\mathrm{E}_{\mathrm{n}}=-\frac{\mathrm{R}_{\mathrm{H}}(1)^{2}}{\mathrm{n}^{2}}$
$-\frac{\mathrm{R}_{\mathrm{H}}}{9}=-\frac{\mathrm{R}_{\mathrm{H}}}{\mathrm{n}^{2}}$
$\therefore \mathrm{n}^{2}=9$
$\mathrm{E}_{\mathrm{n}}=-\frac{\mathrm{R}_{\mathrm{H}} \mathrm{Z}^{2}}{\mathrm{n}^{2}}$
where, $Z=$ atomic number, $n^{2}=$ degeneracy
For $\mathrm{H}$-atom, $\mathrm{E}_{\mathrm{n}}=-\frac{\mathrm{R}_{\mathrm{H}}(1)^{2}}{\mathrm{n}^{2}}$
$-\frac{\mathrm{R}_{\mathrm{H}}}{9}=-\frac{\mathrm{R}_{\mathrm{H}}}{\mathrm{n}^{2}}$
$\therefore \mathrm{n}^{2}=9$
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