The difference between orbital angular momentum of an electron in a 4 f orbital and another electron in a 4 s orbital is

Question

The difference between orbital angular momentum of an electron in a 4 f orbital and another electron in a 4 s orbital is (A) 2 √3 (B) 3 √2 (C) 3 (D) 2

Tardigrade Admin · Accepted Answer

Orbital angular momentum =√l(l+1) (h/2 π) For 4 f orbital; l=3 Orbital angular momentum =√3(3+1) (h/2 π)=2 √3 (h/2 π) For 4 s orbital; l=0 ∴ Orbital angular momentum =0 ∴ Difference =2 √3 (h/2 π)

Q. The difference between orbital angular momentum of an electron in a $4 f$ orbital and another electron in a $4 s$ orbital is

$23$

$32$

$3$

$2$

Orbital angular momentum $= l (l + 1) \frac{h}{2 π}$
For $4 f$ orbital; $l = 3$
Orbital angular momentum
$= 3 (3 + 1) \frac{h}{2 π} = 23 \frac{h}{2 π}$
For $4 s$ orbital; $l = 0$
$∴$ Orbital angular momentum $= 0$
$∴$ Difference $= 23 \frac{h}{2 π}$

Q. The difference between orbital angular momentum of an electron in a 4f orbital and another electron in a 4s orbital is

23​

32​

3

2

Orbital angular momentum =l(l+1)​2πh​ For 4f orbital; l=3 Orbital angular momentum =3(3+1)​2πh​=23​2πh​ For 4s orbital; l=0 ∴ Orbital angular momentum =0 ∴ Difference =23​2πh​

Q. The difference between orbital angular momentum of an electron in a $4 f$ orbital and another electron in a $4 s$ orbital is

$23$

$32$

$3$

$2$

Orbital angular momentum $= l (l + 1) \frac{h}{2 π}$
For $4 f$ orbital; $l = 3$
Orbital angular momentum
$= 3 (3 + 1) \frac{h}{2 π} = 23 \frac{h}{2 π}$
For $4 s$ orbital; $l = 0$
$∴$ Orbital angular momentum $= 0$
$∴$ Difference $= 23 \frac{h}{2 π}$