J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    If the electron in hydrogen atom jumps from the third to second orbit, the wavelength of the emitted radiation in terms of Rydberg constant R is given by :

    A)  \[\lambda =\frac{36}{5R}\]                                        

    B)  \[\lambda =\frac{5R}{36}\]

    C)  \[\lambda =\frac{5}{R}\]                                             

    D)  \[\lambda =\frac{R}{6}\]

    Correct Answer: A

    Solution :

                    From Bohrs model of atom, the wave number is given by \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where R is Rydbergs constant and\[{{n}_{1}}\]and\[{{n}_{2}}\]the energy levels. Given,      \[{{n}_{1}}=2,{{n}_{2}}=3\] \[\therefore \]  \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[\frac{1}{\lambda }=R\left[ \frac{5}{36} \right]\] \[\Rightarrow \]               \[\lambda =\frac{36}{5R}\] This gives the corresponding wavelength of Balmer series.


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