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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Mock Test - Modern Physics

  • question_answer
    Hydrogen (H), deuterium (4), singly ionized helium (\[H{{e}^{+}}\] and doubly ionized lithium (Li) all have one electroi around the nucleus. Consider n = 2 to n = 1 transition the wavelengths of emitted radiations are \[{{\lambda }_{1}},\]\[{{\lambda }_{2}},\]\[{{\lambda }_{3}},\] and \[{{\lambda }_{4}}\] respectively. Then approximately

    A) \[{{\lambda }_{1}}={{\lambda }_{2}}=4{{\lambda }_{3}}=9{{\lambda }_{4}}\]

    B) \[4{{\lambda }_{1}}=2{{\lambda }_{2}}=2{{\lambda }_{3}}={{\lambda }_{4}}\]

    C) \[{{\lambda }_{1}}=2{{\lambda }_{2}}=2\sqrt{2}{{\lambda }_{3}}=3\sqrt{2}{{\lambda }_{4}}\]

    D) \[{{\lambda }_{1}}={{\lambda }_{2}}=2{{\lambda }_{3}}=3\sqrt{2}{{\lambda }_{4}}\]  

    Correct Answer: A

    Solution :

    [a] Using \[\Delta E\propto {{Z}^{2}}\](\[\therefore \]\[{{n}_{1}}\]and \[{{n}_{2}}\] are same) \[\Rightarrow \frac{hc}{\lambda }\propto {{Z}^{2}}\Rightarrow \lambda {{Z}^{2}}=\text{constant}\] \[\Rightarrow {{\lambda }_{1}}{{Z}^{2}}_{1}={{\lambda }_{2}}{{Z}^{2}}_{2}={{\lambda }_{3}}{{Z}^{2}}_{3}={{\lambda }_{4}}{{Z}^{2}}_{4}\] \[\Rightarrow {{\lambda }_{1}}\times 1={{\lambda }_{2}}\times {{1}^{2}}={{\lambda }_{3}}\times {{2}^{2}}={{\lambda }_{4}}\times {{3}^{3}}\] \[\Rightarrow {{\lambda }_{1}}={{\lambda }_{2}}=4{{\lambda }_{3}}=9{{\lambda }_{4}}\]


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