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JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    Hydrogen \[{{(}_{1}}{{H}^{1}}),\]Deuterium\[{{(}_{1}}{{H}^{2}}),\] singly ionised Helium \[{{{{(}_{2}}H{{e}^{4}})}^{+}}\]and doubly ionised lithium \[{{{{(}_{3}}L{{i}^{6}})}^{++}}\]allhave one electron around the nucleus. Consider an electron transition from n = 2 ton = l. If the wave lengths of emitted radiation are \[{{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}}\]and \[{{\lambda }_{4}}\]respectively then approximately whichone of the following is correct ?   JEE Main  Solved  Paper-2014

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

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

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

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

    Correct Answer: A

    Solution :

    \[\frac{1}{\lambda }=\frac{R{{Z}^{2}}}{\left( 1+\frac{me}{M} \right)}\left[ 1-\frac{1}{4} \right]\] \[\frac{1}{\lambda }=\frac{3}{4}{{R}_{0}}.\frac{{{Z}^{2}}}{\left( 1+\frac{me}{M} \right)}\] [M = mass of mucleus] \[\frac{1}{\lambda }\propto \frac{1}{\left( 1+\frac{me}{M} \right)}\,\,\frac{1}{\lambda }\propto \frac{1}{\left( 1+\frac{me}{2{{M}_{p}}} \right)}\] [\[{{\text{M}}_{\text{P}}}=\]mass of H − atom] \[\frac{1}{{{\lambda }_{3}}}\propto \frac{4}{\left( 1+\frac{me}{4{{M}_{p}}} \right)}\,\,\frac{1}{{{\lambda }_{4}}}\propto \frac{9}{\left( 1+\frac{me}{6{{M}_{p}}} \right)}\] If\[\frac{me}{{{M}_{p}}}<<1:\]we have \[{{\lambda }_{1}}={{\lambda }_{2}}=4{{\lambda }_{3}}=9{{\lambda }_{4}}\]


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