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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2024 · 6 April · Evening Shift · Physics · Atomic Physics

The longest wavelength associated with Paschen series is : (Given...

The longest wavelength associated with Paschen series is : (Given RH=1.097×107SI\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI} unit)

Options

  1. A3.646×10−6 m3.646 \times 10^{-6} \mathrm{~m}
  2. B1.876×10−6 m1.876 \times 10^{-6} \mathrm{~m}
  3. C2.973×10−6 m2.973 \times 10^{-6} \mathrm{~m}
  4. D1.094×10−6 m1.094 \times 10^{-6} \mathrm{~m}

Correct answer

B. 1.876×10−6 m1.876 \times 10^{-6} \mathrm{~m}

Step-by-step solution

For longest wavelength in Paschen's series: 1λ=R\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]ForlongestFor longest\mathrm{n}_1=3 \mathrm{n}_2=4 1λ=R[1(3)2−1(4)2]1λ=R[19−116]1λ=R[16−916×9]⇒λ=16×97R=16×97×1.097×107λ=1.876×10−6 m\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}
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