JEE MAIN - Chemistry (2023 - 13th April Evening Shift - No. 17)
The orbital angular momentum of an electron in $$3 \mathrm{~s}$$ orbital is $$\frac{x h}{2 \pi}$$. The value of $$x$$ is ____________ (nearest integer)
Answer
0
Explanation
The orbital angular momentum (L) of an electron can be determined using the formula:
$$L = \sqrt{l(l+1)} \frac{h}{2\pi}$$
Where $$l$$ is the azimuthal quantum number (orbital angular momentum quantum number) and $$h$$ is the Planck's constant.
For a 3s orbital, the principal quantum number (n) is 3 and the azimuthal quantum number (l) is 0, as s orbitals have l=0. Now, let's calculate the orbital angular momentum:
$$L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$$
Thus, the orbital angular momentum of an electron in a 3s orbital is 0. So, the value of $$x$$ is 0.
$$L = \sqrt{l(l+1)} \frac{h}{2\pi}$$
Where $$l$$ is the azimuthal quantum number (orbital angular momentum quantum number) and $$h$$ is the Planck's constant.
For a 3s orbital, the principal quantum number (n) is 3 and the azimuthal quantum number (l) is 0, as s orbitals have l=0. Now, let's calculate the orbital angular momentum:
$$L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$$
Thus, the orbital angular momentum of an electron in a 3s orbital is 0. So, the value of $$x$$ is 0.
Comments (0)
