Solved papers for JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

done JEE Main Online Paper (Held On 23 April 2013) Total Questions - 1

• question_answer1)                 In the Bohr?s model of hydrogen- like atom the force between the nucleus and the electron is modified as                 \[\operatorname{F}=\frac{{{e}^{2}}}{4\pi {{\in }_{0}}}\left( \frac{1}{{{\operatorname{r}}^{2}}}+\frac{\beta }{{{\operatorname{r}}^{3}}} \right),\] Where \[\beta \]is a constant. For this atom, the radius of the \[{{\operatorname{n}}^{\operatorname{th}}}\] orbit in terms of the Bohr radius \[\left( {{a}_{0}}\frac{{{\in }_{0}}{{\operatorname{h}}^{2}}}{\operatorname{m}\pi {{e}^{2}}} \right)\] is     [JEE Main Online Paper ( Held On 23  April 2013 )
  A)
                   \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}\operatorname{n}-\beta \]                     
  done clear
  B)
                                          \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}{{\operatorname{n}}^{2}}+\beta \]
  done clear
  C)
                                          \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}{{\operatorname{n}}^{2}}-\beta \]               
  done clear
  D)
                                          \[{{\operatorname{r}}_{\operatorname{n}}}={{\operatorname{a}}_{0}}\operatorname{n}+\beta \]
  done clear
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