JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    Based on the equation: \[\Delta E=-2.0\times {{10}^{-18}}J\left( \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right)\]the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level n = 2 will be: \[(h=6.625\times {{10}^{-34}}Js,C=3\times {{10}^{8}}m{{s}^{-1}})\]   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) \[1.325\times {{10}^{-7}}m\]

    B) \[1.325\times {{10}^{-10}}m\]

    C) \[2.650\times {{10}^{-7}}m\]     

    D) \[5.300\times {{10}^{-10}}m\]

    Correct Answer: A

    Solution :

                    \[\Delta E=-2.0\times {{10}^{-18}}\times \left( \frac{1}{{{2}^{2}}}-\frac{1}{{{1}^{2}}} \right)\] \[=-2.0\times {{10}^{-18}}\times \frac{-3}{4}\]\[=1.5\times {{10}^{-18}}\] \[\Delta E=\frac{hc}{\lambda }\] \[\lambda =\frac{hc}{\Delta E}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.5\times {{10}^{-18}}}=1.325\times {{10}^{-7}}m\]                


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