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Q.
The degeneracy of the level of hydrogen atoms that contain the energy of $\left(\frac{- R _{ H }}{16}\right)$ is
TS EAMCET 2020
Solution:
Given, $E=\frac{-R_{H}}{16}$
We know that, $E=\frac{-R_{H}}{n^{2}}=\frac{-R_{H}}{16},$
$ \therefore n=4$
Thus, degeneracy for $n=4$ will be,
Hence, total degeneracy of orbital for $n=4$ will be $ 16$ .