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JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    The electron of a hydrogen atom makes a transition from the \[{{(n+1)}^{th}}\]orbit to the \[{{n}^{th}}\] orbit. For large n the wavelength of the emitted radiation is proportional to   JEE Main Online Paper (Held On 07 May 2012)

    A) n                                            

    B)                        \[{{n}^{3}}\]

    C)                        \[{{n}^{4}}\]                                      

    D)                        \[{{n}^{2}}\]

    Correct Answer: B

    Solution :

                    lf\[{{n}_{1}}=n\]and\[{{n}_{2}}=n+1\] Maximum wavelength \[{{\lambda }_{\max }}=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{\left( 2n+1 \right)R}\] Therefore, for large \[n,{{\lambda }_{\max }}\propto {{n}^{3}}\]


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