Atomic Structure Question 376
The orbital angular momentum for an electron revolving in an orbit is given by $ \sqrt{l(l+1)}\hbar $ . This momentum for $ d^{5},,d^{6},,p^{3} $ electron will be respectively
Options:
A) $ \sqrt{6},\frac{h}{2\pi },,\sqrt{7},\frac{h}{2\pi },,\sqrt{2},\frac{h}{2\pi } $
B) $ 0,\frac{\sqrt{6}h}{2\pi },,\frac{\sqrt{2}h}{2\pi } $
C) $ \sqrt{6},\frac{h}{2\pi },,\sqrt{6}\frac{6}{2\pi },,\sqrt{2}\frac{h}{2\pi } $
D) $ \sqrt{6},\frac{h}{2\pi },,\sqrt{6}\frac{h}{2\pi },,\sqrt{3}\frac{h}{2\pi } $
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Answer:
Correct Answer: C
Solution:
- For electron $ L=\sqrt{2(2+1)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi } $ For electron $ L=\sqrt{1(1+1)}\frac{h}{2\pi }=\sqrt{2}\frac{h}{2\pi } $
BETA
