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 \[{{\lambda }_{1}}={{\lambda }_{2}}=4{{\lambda }_{3}}=9{{\lambda }_{4}}\]You need to login to perform this action.
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