Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is
  1. $\frac{5}{27}$
  2. $\frac{3}{23}$
  3. $\frac{7}{29}$
  4. $\frac{9}{31}$

Solution

Wavelength for Lyman series $\lambda_L=\frac{1}{R\left(1-\frac{1}{4}\right)}=\frac{4}{3 R}$ and wavelength for Balmer series $\begin{aligned} & \lambda_B=\frac{1}{R\left(\frac{1}{4}-\frac{1}{9}\right)}=\frac{1}{R\left(\frac{5}{36}\right)}=\frac{36}{5 R} \\ & \therefore \quad \frac{\lambda_L}{\lambda_B}=\frac{4}{3 R} \times \frac{5 R}{36}=\frac{5}{27} \\ & \Rightarrow \quad \lambda_L: \lambda_B=5: 27 \\ & \end{aligned}$

Asked in: NEET 2013 (All India)

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