A) \[121.6\text{ }nm\]
B) \[182.4\text{ }nm\]
C) \[243.4\text{ }nm\]
D) \[364.8\text{ }nm\]
Correct Answer: A
Solution :
The wavelength (X) of lines is given by \[\frac{1}{\lambda }=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right)\] For Lyman series, the shortest wavelength is for \[n=\infty \] and longest is for \[n=2\]. \[\therefore \] \[\frac{1}{{{\lambda }_{S}}}=R\left( \frac{1}{{{1}^{2}}} \right)\] ?..(i) \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{1}-\frac{1}{{{2}^{2}}} \right)=\frac{3}{4}R\] ??(ii) Dividing Eq. (ii) by Eq. (i) we; get \[\frac{{{\lambda }_{L}}}{{{\lambda }_{S}}}=\frac{4}{3}\] Given, \[{{\lambda }_{S}}=91.2nm\] \[\Rightarrow \] \[{{\lambda }_{L}}=91.2\times \frac{4}{3}=121.6nm\]
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