The degeneracy of the energy level of a hydro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(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

Similar Questions

Number of completely filled orbitals in xenon atom $(Xe)$ is

The shape of $p$-orbital is

How many unpaired electrons are present in a carbon atom in its ground state?

The maximum number of unpaired electrons that can be present in $d$-orbitals is

Vedclass Test Series

Exam Paper Generator

Online Exam Module