What is the degeneracy of the level of H-atom that has energy (-(RH/9)) ?

Question

What is the degeneracy of the level of H-atom that has energy (-(RH/9)) ? (A) 16 (B) 9 (C) 4 (D) 1

Tardigrade Admin · Accepted Answer

Energy of the electron in the nth orbit in terms of R H is En=-(R H Z2/n2) where, Z= atomic number, n2= degeneracy For H-atom, En=-(R H (1)2/n2) -(R H /9) =-(R H /n2) ∴ n2 =9

Q. What is the degeneracy of the level of $H$ -atom that has energy $(- \frac{R _{H}}{9}) ?$

16

9

4

1

Energy of the electron in the $n$ th orbit in terms of $R_{H}$ is
$E_{n} = - \frac{R _{H} Z ^{2}}{n ^{2}}$
where, $Z =$ atomic number, $n^{2} =$ degeneracy
For H-atom, $E_{n} = - \frac{R _{H} ( 1 ) ^{2}}{n ^{2}}$
$- \frac{R _{H}}{9} = - \frac{R _{H}}{n ^{2}}$
$∴ n^{2} = 9$

Q. What is the degeneracy of the level of H-atom that has energy (−9RH​​)?

16

9

4

1

Energy of the electron in the nth orbit in terms of RH​ is En​=−n2RH​Z2​ where, Z= atomic number, n2= degeneracy For H-atom, En​=−n2RH​(1)2​ −9RH​​=−n2RH​​ ∴n2=9

Q. What is the degeneracy of the level of $H$ -atom that has energy $(- \frac{R _{H}}{9}) ?$

Energy of the electron in the $n$ th orbit in terms of $R_{H}$ is
$E_{n} = - \frac{R _{H} Z ^{2}}{n ^{2}}$
where, $Z =$ atomic number, $n^{2} =$ degeneracy
For H-atom, $E_{n} = - \frac{R _{H} ( 1 ) ^{2}}{n ^{2}}$
$- \frac{R _{H}}{9} = - \frac{R _{H}}{n ^{2}}$
$∴ n^{2} = 9$