Question

the ratio of Orbital angular momentum and spin angular momentum of an electron in p orbital is

Answer 1

Answer:2×root2 divided by root 3

Explanation:

Answer 2

Answer:

An electron in a p-orbital has an angular momentum ratio of  $\frac{2\sqrt{2} }{\sqrt{3} }$ between its orbital and spin angular momentum.

Explanation:

We know that for an electron in the p-orbital,

$l=1\\s=\frac{1}{2}$

By taking ratios :

$\frac{L}{S} =\frac{\sqrt{l(l+1)h} }{\sqrt{s(s+1)h} }$

$=\frac{\sqrt{2} }{\frac{\sqrt{3} }{2} }$

Now, we bring the denominator of the denominator to the numerator:

$=\frac{2\sqrt{2} }{\sqrt{3} }$

Hence, the ratio of the orbital angular momentum with the spin angular momentum of an electron in the p-orbital is : $\frac{2\sqrt{2} }{\sqrt{3} }$

Orbital angular momentum:

Free-floating electrons have the ability to project quantized orbital angular momentum in the direction of propagation. A phase proportionate to the azimuthal angle or helical wavefronts, respectively, corresponds to this orbital angular momentum.

Spin angular momentum:

Atomic nuclei and composite particles (hadrons), as well as elementary particles, carry spin, an intrinsic type of angular momentum.