RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer The wavelength of light emitted due to transition of electron from second orbit to first orbit in hydrogen atom is

    A)  \[6563\overset{\text{o}}{\mathop{\text{A}}}\,\]                           

    B)  \[4102\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)  \[4861\overset{\text{o}}{\mathop{\text{A}}}\,\]           

    D)         \[1215\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: D

    Solution :

    From Bohrs theory, wavelength of radiations emitted in H-atom \[\frac{1}{\lambda }=R\left( \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right)\] Here \[{{n}_{1}}=2,1,R=1.097\times {{10}^{7}}{{m}^{-1}}\] \[\therefore \]  \[\frac{1}{\lambda }=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3}{4}R\] \[\Rightarrow \]               \[\frac{1}{\lambda }=\frac{4}{3R}=\frac{4}{3\times 1.097\times {{10}^{7}}}\]                                 \[\begin{align}   & =1.215\times {{10}^{-7}}m \\  & =1215\,\overset{\text{0}}{\mathop{\text{A}}}\, \\ \end{align}\]


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