A) \[1.11\times {{10}^{34}}J\ \sec \]
B) \[1.51\times {{10}^{-31}}J\ \sec \]
C) \[2.11\times {{10}^{-34}}J\ \sec \]
D) \[3.72\times {{10}^{-34}}J\ \sec \]
Correct Answer: C
Solution :
The electron is in the second orbit (n=2) Hence \[L=\frac{nh}{2\pi }=\frac{2h}{2\pi }=\frac{6.6\times {{10}^{-34}}}{\pi }\]\[=2.11\times {{10}^{-34}}J\text{-}sec\]
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