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JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

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    Taking Rydberg?s constant \[{{R}_{H}}=1.097\times {{10}^{7}}m\] first and second wavelength of Balmer series in hydrogen spectrum is [Pb. PMT 2004]

    A)            2000 Å, 3000 Å                     

    B)            1575 Å, 2960 Å

    C)            6529 Å, 4280 Å                     

    D)            6552 Å, 4863 Å

    Correct Answer: D

    Solution :

               \[\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right].\] For first wavelength, \[{{n}_{1}}=2\], \[{{n}_{2}}=3\] Þ\[{{\lambda }_{1}}=6563\,{\AA}\]. For second wavelength, \[{{n}_{1}}=2\], \[{{n}_{2}}=4\] Þ \[{{\lambda }_{2}}=4861\,{\AA}\]


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