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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    If an electron has orbital angular momentum quantum number\[l=7,\]then it will have an orbital angular momentum equal to

    A)  \[7\left( \frac{h}{2\pi } \right)\]                               

    B)  \[42\left( \frac{h}{2\pi } \right)\]

    C)   \[\sqrt{7}\left( \frac{h}{2\pi } \right)\]                

    D)  \[\sqrt{56}\left( \frac{h}{2\pi } \right)\]

    Correct Answer: D

    Solution :

                     \[L=\sqrt{l(l+1)}\frac{n}{2\pi }\] \[L=\sqrt{7(7+1)}\frac{h}{2\pi }\] \[L=\sqrt{7\times 8}\frac{h}{2\pi }=\sqrt{56}\left( \frac{h}{2\pi } \right)\]


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