NEET Sample Paper NEET Sample Test Paper-18

  • question_answer In a hydrogen atom, the binding energy of the electron in the ground state is \[{{E}_{1}}.\]Then the frequency of revolution of the electron in the nth orbit is

    A) \[\frac{2{{E}_{1}}}{{{n}^{3}}h}\]                                  

    B) \[\frac{2{{E}_{1}}{{n}^{3}}}{h}\]

    C) \[\sqrt{\frac{2m{{E}_{1}}}{{{n}^{3}}h}}\]                    

    D) \[\frac{{{E}_{1}}{{n}^{2}}}{h}\]   

    Correct Answer: A

    Solution :

    Binding energy of electron in the nth orbit is \[{{E}_{n}}={{E}_{1}}/{{n}^{2}}.\] Now, \[m{{v}_{n}}{{r}_{n}}=\frac{nh}{2\pi }\]and \[\frac{1}{2}mv_{n}^{2}={{E}_{n}}\] \[\Rightarrow \]            \[\frac{\left( \frac{1}{2} \right)mv_{n}^{2}}{m{{v}_{n}}{{r}_{n}}}=\frac{{{E}_{n}}}{\frac{nh}{2\pi }}\] \[\Rightarrow \]\[\frac{{{v}_{n}}}{{{r}_{n}}}=\frac{4\pi {{E}_{n}}}{nh}=\frac{4\pi {{E}_{1}}}{{{n}^{3}}h}\Rightarrow \frac{4\pi {{E}_{1}}}{{{n}^{3}}h}\] \[\therefore \]    \[{{f}_{n}}=\frac{{{\omega }_{n}}}{2\pi }=\frac{2{{E}_{1}}}{{{n}^{3}}h}\] Hence, the correction option is [a].

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