The degeneracy of the level of hydrogen atoms that contain the energy of $\left(\frac{-R_{\mathrm{H}}}{16}\right)$ is

• A

  4
• Correct answer
  B

  16
• C

  9
• D

  12

Solution

Given, $E=\frac{-R_{\mathrm{H}}}{16}$ We know that, $E=\frac{-R_{\mathrm{H}}}{n^2}=\frac{-R_{\mathrm{H}}}{16}, \therefore n=4$ Thus, degeneracy for $n=4$ will be,
Hence, total degeneracy of orbital for $n=4$ will be 16 .

Hence, the correct answer is option: 'B'.