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

  • question_answer
    If the series limit wavelength of Lyman series for the hydrogen atom is \[912\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\], then the series limit wavelength for Balmer series of hydrogen atoms is

    A)  \[912\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B)  \[912\text{ }\times \text{2}\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)  \[912\text{ }\times 4\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D)  \[\frac{912}{2}\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

    [c]  \[\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] For limiting wavelength of Lyman series \[{{n}_{1}}=1,\,\,\,{{n}_{2}}=\infty \frac{1}{{{\lambda }_{L}}=R}\] For limiting wavelength of Balmer series \[{{n}_{1}}=1,{{n}_{2}}=\infty \] \[\frac{1}{{{\lambda }_{B}}}=R\left( \frac{1}{4} \right)\Rightarrow {{\lambda }_{B}}=\frac{4}{R}\] \[\therefore {{\lambda }_{B}}=4{{\lambda }_{1}}=4\times 912\overset{\text{o}}{\mathop{\text{A}}}\,.\]


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