The magnitude of the orbital angular momentum of an electron is given by L - -5 h- How many orbitals of this kind are possible for the subshell-

Orbital angular momentum = √(l(l+1))h2π where, 'l' is the Azimuthal quantum number. Given: nbsp; √(l(l+1))h2π=√(5)hπ ∴ l=4 So, number of possible orbitals = ...

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Standard XI

Chemistry

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?

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Solution

The correct option is D 9
Orbital angular momentum = $\sqrt{l (l + 1)} \frac{h}{2 π}$
where, ' $l$ ' is the Azimuthal quantum number.

Given:
$\sqrt{l (l + 1)} \frac{h}{2 π} = \sqrt{5} \frac{h}{π}$
$∴ l = 4$
So, number of possible orbitals = $2 l + 1 = 9$

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The magnitude of the orbital angular momentum of an electron is given by L=√5hπ. How many orbitals of this kind are possible for the subshell?

The correct option is D 9Orbital angular momentum = √l(l+1)h2π where, 'l' is the Azimuthal quantum number. Given: √l(l+1)h2π=√5hπ ∴l=4 So, number of possible orbitals = 2l+1=9

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?

The correct option is D 9
Orbital angular momentum = $\sqrt{l (l + 1)} \frac{h}{2 π}$
where, ' $l$ ' is the Azimuthal quantum number.

Given:
$\sqrt{l (l + 1)} \frac{h}{2 π} = \sqrt{5} \frac{h}{π}$
$∴ l = 4$
So, number of possible orbitals = $2 l + 1 = 9$