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 (integer)
0
Solution
The orbital angular momentum (L) of an electron can be determined using the formula:
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$L = \sqrt{l(l+1)} \frac{h}{2\pi}$
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Where $l$ is the azimuthal quantum number (orbital angular momentum quantum number) and $h$ is the Planck's constant.
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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:
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$L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$
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Thus, the orbital angular momentum of an electron in a 3s orbital is 0. So, the value of $x$ is 0.
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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