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JEE Main & Advanced AIEEE Solved Paper-2013

  • question_answer
    Energy of an electron is given by\[E=-2.178\times \]\[{{10}^{18}}J\left( \frac{{{Z}^{2}}}{{{n}^{2}}} \right)\]. Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be: (\[h=6.62\times {{10}^{-34}}Js\]and\[c=3.0\times {{10}^{-8}}m{{s}^{-1}}\])     AIEEE Solevd Paper-2013

    A) \[1.214\times {{10}^{-7}}m\]     

    B) \[2.816\times {{10}^{-7}}m\]

    C)        \[6.500\times {{10}^{-7}}m\]

    D)        \[8.500\times {{10}^{-7}}m\]

    Correct Answer: A

    Solution :

    \[\frac{1}{\lambda }={{R}_{H}}{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[Z=1\] \[{{n}_{1}}=1\] \[{{n}_{2}}=2\] \[\lambda =1.214\times {{10}^{-7}}m\]


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