Orbital angular momentum of an electron in a particular subshell is √5 (h/π) . Then find the maximum number of electrons which may be present in this subshell.

Question

Orbital angular momentum of an electron in a particular subshell is √5 (h/π) . Then find the maximum number of electrons which may be present in this subshell. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Orbital angular momentum =√l(l+1) (h/2 π) ⇒ √l(l+1) (h/2 π)=√5 (h/π) √l(l+1)=2 √5=√20 ⇒ l=4 . Hence, maximum number of electrons in this subshell = 2(2 l+1)=18.

Q. Orbital angular momentum of an electron in a particular subshell is $5 \frac{h}{π} .$ Then find the maximum number of electrons which may be present in this subshell.

Orbital angular momentum $= l (l + 1) \frac{h}{2 π}$
$\Rightarrow l (l + 1) \frac{h}{2 π} = 5 \frac{h}{π}$
$l (l + 1) = 25 = 20$
$\Rightarrow l = 4.$
Hence, maximum number of electrons in this subshell
$= 2 (2 l + 1) = 18$ .

Q. Orbital angular momentum of an electron in a particular subshell is 5​πh​. Then find the maximum number of electrons which may be present in this subshell.

Orbital angular momentum =l(l+1)​2πh​ ⇒l(l+1)​2πh​=5​πh​ l(l+1)​=25​=20​ ⇒l=4. Hence, maximum number of electrons in this subshell =2(2l+1)=18.

Q. Orbital angular momentum of an electron in a particular subshell is $5 \frac{h}{π} .$ Then find the maximum number of electrons which may be present in this subshell.

Orbital angular momentum $= l (l + 1) \frac{h}{2 π}$
$\Rightarrow l (l + 1) \frac{h}{2 π} = 5 \frac{h}{π}$
$l (l + 1) = 25 = 20$
$\Rightarrow l = 4.$
Hence, maximum number of electrons in this subshell
$= 2 (2 l + 1) = 18$ .