MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Quantum Numbers

The magnitude...

Question

The magnitude of the orbital angular momentum of an electron is given by $L = \frac{\sqrt{5} h}{π}$ . How many orbitals of this kind are possible for the subshell?

A

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 9
Orbital angular momentum = $\sqrt{l (l + 1)} \frac{h}{2 π} w h e r e^{'} l^{'}$ is Azimuthal quantum number.
Given, $\sqrt{l (l + 1)} \frac{h}{2 π} = \sqrt{5} \frac{h}{π}$
$∴ l = 4$
So, number of possible orbitals = 2 $l$ + 1 = 9

Suggest Corrections

thumbs-up

3

Similar questions

Q. The magnitude of the orbital angular momentum of an electron is given by

L = \frac{\sqrt{5} h}{π}

. How many orbitals of this kind are possible for the subshell?

Q. Orbital angular momentum of an electron in a particular subshell is

\sqrt{5} \frac{h}{π}

. Then find the maximum number of electrons which may be present in this subshell.

Q. An electron has an orbital angular momentum equal to

\sqrt{6} \frac{h}{2 π}

. How many z-components of this are possible?

Q. How many degenerate orbitals are present in a subshell if the electron associated with it has an orbital angular momentum

= \frac{2 \sqrt{3} h}{2 π}

Q. Orbital angular momentum of an electron is

\sqrt{3} \frac{h}{π}

. Then, the possible number of orientations of this orbital in space are:

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Quantum Numbers

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon