For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively, are : [2025]

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Q.
For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively, are : [2025]

1 $\sqrt{2} \frac{h}{2 π}$ and $0$

2 $\frac{h}{2 π}$ and $\sqrt{2} \frac{h}{2 π}$

3 $0$ and $\sqrt{6} \frac{h}{2 π}$

4 $0$ and $\sqrt{2} \frac{h}{2 π}$

Ans.
(4)

Orbital Azimuthal quantum number $(l)$ Orbital angular momentum $= \sqrt{l (l + 1)} \frac{h}{2 π}$

$2 s$ 0 $\sqrt{0 (0 + 1)} \frac{h}{2 π} = 0$

$2 p$ 1 $\sqrt{1 (1 + 1)} \frac{h}{2 π} = \sqrt{2} \frac{h}{2 π}$

Solution

Q.
For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively, are : [2025]

1 $\sqrt{2} \frac{h}{2 π}$ and $0$

2 $\frac{h}{2 π}$ and $\sqrt{2} \frac{h}{2 π}$

3 $0$ and $\sqrt{6} \frac{h}{2 π}$

4 $0$ and $\sqrt{2} \frac{h}{2 π}$

Ans.
(4)

Orbital Azimuthal quantum number $(l)$ Orbital angular momentum $= \sqrt{l (l + 1)} \frac{h}{2 π}$

$2 s$ 0 $\sqrt{0 (0 + 1)} \frac{h}{2 π} = 0$

$2 p$ 1 $\sqrt{1 (1 + 1)} \frac{h}{2 π} = \sqrt{2} \frac{h}{2 π}$

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Orbital	Azimuthal quantum number $(l)$	Orbital angular momentum $= \sqrt{l (l + 1)} \frac{h}{2 π}$
$2 s$	0	$\sqrt{0 (0 + 1)} \frac{h}{2 π} = 0$
$2 p$	1	$\sqrt{1 (1 + 1)} \frac{h}{2 π} = \sqrt{2} \frac{h}{2 π}$

Solution

Q. For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively, are : [2025]

1 2 h2π and 0

2 h2π and 2 h2π

3 0 and 6 h2π

4 0 and 2 h2π

Ans. (4) Orbital Azimuthal quantum number (l) Orbital angular momentum =l(l+1) h2π 2s 0 0(0+1) h2π=0 2p 1 1(1+1) h2π=2 h2π

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Q.
For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively, are : [2025]

1 $\sqrt{2} \frac{h}{2 π}$ and $0$

2 $\frac{h}{2 π}$ and $\sqrt{2} \frac{h}{2 π}$

3 $0$ and $\sqrt{6} \frac{h}{2 π}$

4 $0$ and $\sqrt{2} \frac{h}{2 π}$

Ans.
(4)

Orbital Azimuthal quantum number $(l)$ Orbital angular momentum $= \sqrt{l (l + 1)} \frac{h}{2 π}$

$2 s$ 0 $\sqrt{0 (0 + 1)} \frac{h}{2 π} = 0$

$2 p$ 1 $\sqrt{1 (1 + 1)} \frac{h}{2 π} = \sqrt{2} \frac{h}{2 π}$