SATHEE: Atomic Structure Question 492

Atomic Structure Question 492

Question: The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt{l (l + 1)} \frac{h}{2 π}$ . This momentum for $d^{5},, d^{6},, p^{3}$ electron will be respectively

Options:

A) $\sqrt{6}, \frac{h}{2 π},, \sqrt{7}, \frac{h}{2 π},, \sqrt{2}, \frac{h}{2 π}$

B) $0, \frac{\sqrt{6} h}{2 π},, \frac{\sqrt{2} h}{2 π}$

C) $\sqrt{6}, \frac{h}{2 π},, \sqrt{6} \frac{6}{2 π},, \sqrt{2} \frac{h}{2 π}$

D) $\sqrt{6}, \frac{h}{2 π},, \sqrt{6} \frac{h}{2 π},, \sqrt{3} \frac{h}{2 π}$

Show Answer

Answer:

Correct Answer: C

Solution:

Forelectron $L = \sqrt{2 (2 + 1)} \frac{h}{2 π} = \sqrt{6} \frac{h}{2 π}$ Forelectron $L = \sqrt{1 (1 + 1)} \frac{h}{2 π} = \sqrt{2} . \frac{h}{2 π}$

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Atomic Structure Question 492

Question: The orbital angular momentum for an electron revolving in an orbit is given by l(l+1)h2π . This momentum for d5,,d6,,p3 electron will be respectively

Options:

Answer:

Solution:

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Question: The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt{l (l + 1)} \frac{h}{2 π}$ . This momentum for $d^{5},, d^{6},, p^{3}$ electron will be respectively