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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, \[{{\lambda }_{1}}/{{\lambda }_{2}},\]of the photons emitted in this process is : [JEE Main 12-4-2019 Afternoon]

    A) 9/7                  

    B)   7/5

    C) 27/5    

    D)   20/7

    Correct Answer: D

    Solution :

    \[\frac{1}{\lambda }=R\left( \frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} \right)\] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{7}{9\times 16} \right)\] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[=R\left( \frac{5}{4\times 9} \right)\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{\frac{5}{36}}{\frac{7}{9\times 16}}\]\[=\frac{20}{7}\]


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