NEET Sample Paper NEET Sample Test Paper-19

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

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

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

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

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

    Correct Answer: C

    Solution :

    For series limit of Balmer series. \[{{n}_{i}}=\infty ,{{n}_{f}}=2\] \[\frac{l}{{{\lambda }_{b}}}=R\left[ \frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} \right]\]or \[\frac{1}{{{\lambda }_{b}}}=R\left[ \frac{1}{4}-\frac{1}{{{\infty }^{2}}} \right]=\frac{R}{4}\] \[\Rightarrow \]\[{{\lambda }_{b}}=\frac{4}{R}.\] where\[{{\lambda }_{b}}\] is the series limit of the Balmer series. Limit of Lymen series. \[\frac{1}{{{\lambda }_{\ell }}}=R\left[ 1-\frac{1}{\infty } \right]\] Where \[{{\lambda }_{\ell }}\]is the series limit of lymen series. \[\therefore \]\[{{\lambda }_{\ell }}=\frac{1}{R}\] So, \[\frac{{{\lambda }_{b}}}{{{\lambda }_{\ell }}}=4\Rightarrow {{\lambda }_{0}}={{\lambda }_{\ell }}\times 4=912\times 4\overset{o}{\mathop{A}}\,\]             Hence, the correction option is [c].


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