The longest wavelength associated with Paschen series is : (Given $\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7…

The longest wavelength associated with Paschen series is : (Given $\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI}$ unit)
  1. $3.646 \times 10^{-6} \mathrm{~m}$
  2. $1.876 \times 10^{-6} \mathrm{~m}$
  3. $2.973 \times 10^{-6} \mathrm{~m}$
  4. $1.094 \times 10^{-6} \mathrm{~m}$

Solution

For longest wavelength in Paschen's series: $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$ For longest $\mathrm{n}_1=3$ $\mathrm{n}_2=4$ $\begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{16-9}{16 \times 9}\right] \\ & \Rightarrow \lambda=\frac{16 \times 9}{7 \mathrm{R}}=\frac{16 \times 9}{7 \times 1.097 \times 10^7} \\ & \lambda=1.876 \times 10^{-6} \mathrm{~m}\end{aligned}$

Asked in: JEE Main 2024 (06 Apr Shift 2)

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