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Q.
Taking Rydbergs constant $ {{R}_{H}}=1.097 $ $ \times {{10}^{7}}m $ first and second wavelength of Balmer series m hydrogen spectrm, is:
ManipalManipal 2000
Solution:
The wavelength of the lines in Balmer series is represented by, $ \frac{1}{\lambda }={{R}_{H}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{n_{2}^{2}} \right] $ For first wavelength $ \frac{1}{{{\lambda }_{1}}}=1.097\times {{10}^{7}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right] $ $ =1.524\times {{10}^{6}}m $ Or $ {{\lambda }_{1}}=\frac{1}{1.524\times {{10}^{6}}}=6.562\times {{10}^{-7}} $ Or $ {{\lambda }_{1}}=6562\overset{o}{\mathop{\text{A}}}\, $ For second wavelength $ \frac{1}{{{\lambda }_{1}}}=1.097\times {{10}^{-7}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right] $ $ =2.056\times {{10}^{6}} $ Or $ {{\lambda }_{2}}=\frac{1}{2.056\times {{10}^{6}}} $ Or $ {{\lambda }_{2}}=4.863-7\times 10\times {{10}^{-10}} $ Or $ {{\lambda }_{2}}=4863\overset{o}{\mathop{\text{A}}}\, $