The total number of orbitals in the principal shell of He + that has energy equal to (- Rhc /4) (where R is Rydberg's constant)

Question

The total number of orbitals in the principal shell of He + that has energy equal to (- Rhc /4) (where R is Rydberg's constant) (A) 4 (B) 8 (C) 16 (D) 32

Tardigrade Admin · Accepted Answer

T.E. =-Rhc × (z2/n2) (- Rhc /4)=- Rhc × (4/ n 2) n2=16 ⇒ n=4

Q. The total number of orbitals in the principal shell of $H e^{+}$ that has energy equal to $\frac{- R h c}{4}$ (where $R$ is Rydberg's constant)

4

8

16

32

T.E. $= - R h c \times \frac{z ^{2}}{n ^{2}}$

$\frac{- R h c}{4} = - R h c \times \frac{4}{n ^{2}}$

$n^{2} = 16$

$\Rightarrow n = 4$

Q. The total number of orbitals in the principal shell of He+ that has energy equal to 4−Rhc​ (where R is Rydberg's constant)

4

8

16

32

T.E. =−Rhc×n2z2​ 4−Rhc​=−Rhc×n24​ n2=16 ⇒n=4

Q. The total number of orbitals in the principal shell of $H e^{+}$ that has energy equal to $\frac{- R h c}{4}$ (where $R$ is Rydberg's constant)

T.E. $= - R h c \times \frac{z ^{2}}{n ^{2}}$

$\frac{- R h c}{4} = - R h c \times \frac{4}{n ^{2}}$

$n^{2} = 16$

$\Rightarrow n = 4$