The degeneracy of the level of hydrogen atoms that contain the energy of ((- R H /16)) is

Question

The degeneracy of the level of hydrogen atoms that contain the energy of ((- R H /16)) is (A) 4 (B) 16 (C) 9 (D) 12

Tardigrade Admin · Accepted Answer

Given, E=(-RH/16) We know that, E=(-RH/n2)=(-RH/16), ∴ n=4 Thus, degeneracy for n=4 will be, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/1512622121c86f9784595a1d73abf549-.png" /> Hence, total degeneracy of orbital for n=4 will be 16 .

Q. The degeneracy of the level of hydrogen atoms that contain the energy of $(\frac{- R _{H}}{16})$ is

4

16

9

12

Given, $E = \frac{- R _{H}}{16}$
We know that, $E = \frac{- R _{H}}{n ^{2}} = \frac{- R _{H}}{16},$
$∴ n = 4$
Thus, degeneracy for $n = 4$ will be,

Hence, total degeneracy of orbital for $n = 4$ will be $16$ .

Q. The degeneracy of the level of hydrogen atoms that contain the energy of (16−RH​​) is

4

16

9

12

Given, E=16−RH​​ We know that, E=n2−RH​​=16−RH​​, ∴n=4 Thus, degeneracy for n=4 will be, Hence, total degeneracy of orbital for n=4 will be 16 .

Q. The degeneracy of the level of hydrogen atoms that contain the energy of $(\frac{- R _{H}}{16})$ is

Given, $E = \frac{- R _{H}}{16}$
We know that, $E = \frac{- R _{H}}{n ^{2}} = \frac{- R _{H}}{16},$
$∴ n = 4$
Thus, degeneracy for $n = 4$ will be,

Hence, total degeneracy of orbital for $n = 4$ will be $16$ .