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

  • question_answer
    Taking Rydberg's constant \[R=1.097\times {{10}^{7}}m\], first and second wavelength of Balmer series in hydrogen spectrum is

    A)  \[2000\text{ }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,,\text{ }3000\text{ }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,~\]

    B)  \[\text{1575 }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,\text{, }\!\!~\!\!\text{ 2960 }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,\]

    C)  \[\text{6529 }\overset{\text{o}}{\mathop{\text{A}}}\,\text{, }\,\text{4280 }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,\]

    D)  \[6552\text{ }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,,\text{ }4863\text{ }\overset{\text{o}}{\mathop{\text{A }\!\!~\!\!\text{ }}}\,~~\]

    Correct Answer: D

    Solution :

    [d] \[\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right].\] For first wavelength, \[{{n}_{1}}\] \[=2,{{n}_{2}}=3\] \[\Rightarrow {{\lambda }_{1}}=6563\overset{\text{o}}{\mathop{\text{A}}}\,.\] For second wavelength, n, \[=2,{{n}_{2}}=4\] \[\Rightarrow {{\lambda }_{2}}=4861\overset{\text{o}}{\mathop{\text{A}}}\,\]


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