BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    In hydrogen atom, the electron is moving round the nucleus with velocity \[2.18\,\times {{10}^{6\,}}m/s\] in an orbit of radius \[0.528\,\overset{\text{o}}{\mathop{\text{A}}}\,\]. The acceleration of the electron is:            

    A)  \[9\times {{10}^{18}}m/{{s}^{2}}\]                                      

    B)         \[9\times {{10}^{22}}m/{{s}^{2}}\]

    C)         \[9\times {{10}^{-22}}m/{{s}^{2}}\]            

    D)         \[9\times {{10}^{12}}m/{{s}^{2}}\]

    Correct Answer: B

    Solution :

    Acceleration of electron moving round the nucleus is \[a=\frac{{{v}^{2}}}{r}=\frac{{{(2.18\times {{10}^{6}})}^{2}}}{0.528\times {{10}^{-10}}}\] Given, \[v=2.18\times {{10}^{6}}m/s,\text{ }r=0.528\times {{10}^{-10}}m\] Substituting the values in the relation, we have \[a=\frac{{{(2.18\times {{10}^{6}})}^{2}}}{0.528\times {{10}^{-10}}}\] \[\approx \,9\times {{10}^{22}}\,m/{{s}^{2}}\]


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