Manipal Engineering Manipal Engineering Solved Paper-2011

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

    A)  v                                            

    B) \[\frac{22}{5}v\]

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

    D)        \[\frac{2}{5}v\]

    Correct Answer: D

    Solution :

    Velocity of electron in nth orbit                 \[{{v}_{n}}=\frac{Z{{e}^{2}}}{2{{\varepsilon }_{0}}nh}\] \[\Rightarrow \]               \[v\propto \frac{1}{n}\] \[{{v}_{1}}=v,\,\,{{n}_{1}}=2,\,\,{{n}_{2}}=5\] \[\therefore \]  \[\frac{v}{{{v}_{2}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\] \[\therefore \]  \[\frac{v}{{{v}_{2}}}=\frac{5}{2}\] \[\Rightarrow \]               \[{{v}_{2}}=\frac{2v}{5}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner