The degeneracy of the energy level of a hydro | vedclass

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(B) Given, $E = \left( \frac{-R_{H}}{n^2} \right) = \left( \frac{-R_{H}}{16} \right)$.Comparing the denominators, we get $n^2 = 16$, which implies $n

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$16$

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(B) Given, $E = \left( \frac{-R_{H}}{n^2} \right) = \left( \frac{-R_{H}}{16} \right)$.Comparing the denominators, we get $n^2 = 16$, which implies $n = 4$.For a hydrogen atom, the energy depends only on the principal quantum number $n$.The degeneracy of a level with principal quantum number $n$ is given by the formula $n^2$.For $n = 4$, the degeneracy is $4^2 = 16$.This corresponds to the $16$ orbitals present in the $n = 4$ shell: $4s$ ($1$ orbital), $4p$ ($3$ orbitals), $4d$ ($5$ orbitals), and $4f$ ($7$ orbitals), totaling $1 + 3 + 5 + 7 = 16$.

The degeneracy of the energy level of a hydrogen atom that has an energy of $\left(\frac{-R_{H}}{16}\right)$ is

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