A) 121.6 nm
B) 182.4 nm
C) 243.4 nm
D) 364.8 nm
Correct Answer: A
Solution :
The wavelength\[(\lambda )\]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, \[{{\lambda }_{S}}=91.2nm\] \[\Rightarrow \] \[{{\lambda }_{L}}=91.2\times \frac{4}{3}=121.6\,nm\]
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