Question

If $ {{\lambda }_{1}} $ and $ {{\lambda }_{2}} $ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $ {{\lambda }_{1}}:{{\lambda }_{2}} $ is

• A. 1 : 3
• B. 1 : 30
• C. 7 : 50
• D. 7 : 108

Answer 1

Answer: (D)

For first line of Lyman series, $ {{n}_{1}}=1 $ and $ {{n}_{2}}=2 $ $ \therefore $ $ \frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=R\left( 1-\frac{1}{4} \right)=\frac{3R}{4} $ For first line of Paschen series $ {{n}_{1}}=3 $ and $ {{n}_{2}}=4 $ $ \therefore $ $ \frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)=R\left( \frac{1}{9}-\frac{1}{16} \right)=\frac{7R}{144} $ $ \therefore $ $ \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{7R}{144}\times \frac{4}{3R}=\frac{7}{108} $