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Structure of Atom question

2003 · Shift 0 · Q56

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JEE Main
Chemistry
Structure of Atom
2003 · Shift 0 · Q56

Structure of Atom question

2003 · Shift 0 · Q56

JEE MainChemistryStructure of AtomMCQ+4 / −1

The orbital angular momentum for an electron revolving in an orbit is given by

\sqrt {l(l + 1)} {h \over {2\pi }}

. This momentum for an s-electron will be given by

A
zero
B
${h \over {2\pi }}$
C
$\sqrt 2 {h \over {2\pi }}$
D
$+ {1 \over 2}{h \over {2\pi }}$

View written solutionFree

Correct answer: A

The magnitude of orbital angular momentum of an electron is given by

$L = \sqrt{l(l+1)}\,\frac{h}{2\pi}$

where $l$ is the azimuthal quantum number.

For an $s$ -electron, the orbital is an $s$ -orbital, so

$l = 0$

Substitute $l=0$ into the formula:

$L = \sqrt{0(0+1)}\,\frac{h}{2\pi}$

$L = \sqrt{0}\,\frac{h}{2\pi} = 0$

Therefore, the orbital angular momentum for an $s$ -electron is

$0$

Checking options:

A: zero — correct
B: $\dfrac{h}{2\pi}$ — incorrect
C: $\sqrt{2}\dfrac{h}{2\pi}$ — incorrect
D: $+\dfrac{1}{2}\dfrac{h}{2\pi}$ — incorrect

Hence, the correct answer is A.

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