JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    The difference between the longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (atomic number Z) equal to\[\Delta \lambda \]. The value of the Rydberg constant for the given atom is :

    A)  \[\frac{5}{31}\frac{1}{\Delta \lambda .{{Z}^{2}}}\]

    B)  \[\frac{5}{36}\frac{{{Z}^{2}}}{\Delta \lambda .}\]

    C)  \[\frac{31}{5}\frac{1}{\Delta \lambda .{{Z}^{2}}}\]

    D)   none of these

    Correct Answer: C

    Solution :

    [c] \[\Delta \lambda ={{\lambda }_{1}}-{{\lambda }_{2}}\] \[\frac{1}{R\left[ \frac{1}{{{\left( 2 \right)}^{2}}}-\frac{1}{{{\left( 3 \right)}^{2}}} \right]{{Z}^{2}}}=-\frac{1}{R\left[ 1-0 \right]{{Z}^{2}}}\]. On solving, \[R=\frac{31}{5}\frac{1}{\Delta \lambda .{{Z}^{2}}}\]


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