A) \[\Omega \]
B) \[\Omega \]
C) \[\Omega \]
D) none of these
Correct Answer: A
Solution :
The wavelength of spectral line in the Balmer series is given as \[\frac{1}{{{\lambda }_{B}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ??(i) The wavelength of spectral line in Lyman series is given by \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3R}{4}\] ??.(ii) Dividing equation (i) by eqn. (ii), we get \[\frac{1/{{\lambda }_{B}}}{1/{{\lambda }_{L}}}=\frac{5R/36}{3R/4}=\frac{5}{27}\] So, \[{{\lambda }_{L}}=\frac{5}{27}{{\lambda }_{B}}=\frac{5}{27}\times 6563=1215.4\overset{\text{o}}{\mathop{\text{A}}}\,\] From equation (i), we have \[R=\frac{36}{5{{\lambda }_{B}}}=\frac{36}{5\times 6563\times {{10}^{-10}}}\] \[=1.097\times {{10}^{7}}\,{{m}^{-1}}\] \[=1.1\times {{10}^{7}}{{m}^{-1}}\]
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