' $\lambda_1$ ' is the wavelength of series limit of Lyman series, ' $\lambda_2$ ' is the wavelength of the…

' $\lambda_1$ ' is the wavelength of series limit of Lyman series, ' $\lambda_2$ ' is the wavelength of the first line of Lyman series and ' $\lambda_3$ ' is the series limit of the Balmer series. Then the relation between ' $\lambda_1$ ' $\lambda_2$ and $\lambda_3$ is
  1. $\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$
  2. $\frac{1}{\lambda_1}=\frac{1}{\lambda_2}-\frac{1}{\lambda_3}$
  3. $\lambda_2=\lambda_1+\lambda_3$
  4. $\lambda_1=\lambda_2+\lambda_3$

Solution

Series limit of Lyman series is given by $\frac{1}{\lambda_1}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{\infty}\right)=\mathrm{R}$ Series limit of Balmer series given by $\frac{1}{\lambda_3}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{\mathrm{R}}{4}$ First line of Lyman series is given by $\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)=\mathrm{R}-\frac{\mathrm{R}}{4}=\frac{1}{\lambda_1}-\frac{1}{\lambda_3} \\ & \therefore \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3} \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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