NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
                          Consider \[{{3}^{rd}}\] orbit of \[H{{e}^{+}}\] (Helium), using non-relativistic approach, the speed of electron in this orbit will be (given \[K=9\times {{10}^{9}}\] constant, Z = 2 and h (Planck's constant) = \[6.6\times {{10}^{-34}}\text{J}-s\])                        

    A)  \[2.92\times {{10}^{6}}m/s\]                    

    B)  \[1.46\times {{10}^{6}}m/s\]

    C)         \[0.73\times {{10}^{6}}m/s\]   

    D)         \[3.0\times {{10}^{8}}m/s\]

    Correct Answer: B

    Solution :

    Energy of electron in the 3rd orbit of \[H{{e}^{+}}\] is \[{{E}_{3}}=-13.6\times \frac{{{Z}^{2}}}{{{n}^{2}}}eV=-13.6\times \frac{4}{{{3}^{2}}}eV\] \[=-13.6\times \frac{4}{9}\times 1.6\times {{10}^{-19}}J\] From Bohr's model,                             \[{{E}_{3}}=-K{{E}_{3}}=-\frac{1}{2}{{m}_{e}}{{v}^{2}}\] \[\Rightarrow \,\,\,\,\frac{1}{2}\times 9.1\times {{10}^{-31}}\times {{v}^{2}}\]                 \[=-13.6\times \frac{4}{9}\times 1.6\times {{10}^{-19}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{v}^{2}}=\frac{136\times 16\times 4\times 2\times {{10}^{-11}}}{9\times 91}\] or  \[v=1.46\times {{10}^{6}}m/s\]


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