BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    The velocity of an electron in the second orbit of sodium atom (atomic number = 11) is v. The velocity of an electron in its fifth orbit will be

    A)  \[v\]                                    

    B)  \[\frac{22}{5}v\]

    C)  \[\frac{5}{2}v\]                               

    D)  \[\frac{2}{5}v\]

    Correct Answer: D

    Solution :

                     The velocity of nth orbit \[{{v}_{n}}\propto \frac{1}{n}\] \[\frac{{{v}_{5}}}{{{v}_{2}}}=\frac{2}{5}\] Or           \[{{v}_{5}}=\frac{2}{5}{{v}_{2}}=\frac{2}{5}v\]


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