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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2019 · 11 January · Morning Shift · Chemistry · Structure of Atom

Heat treatment of muscular pain involves radiation of wavelength of...

Heat treatment of muscular pain involves radiation of wavelength of about 900nm900 \mathrm{nm}. Which spectral line of H\mathrm{H} atom is suitable for this purpose? [RH=1×105 cm−1⋅h=6.6×10−34Js,c=3×108 ms−1]\left[\mathrm{R}_{\mathrm{H}}=1 \times 10^{5} \mathrm{~cm}^{-1} \cdot \mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]

Options

  1. APaschen, ∞→3\infty \rightarrow 3
  2. BPaschen, 5→35 \rightarrow 3
  3. CBalmer, ∞→2\infty \rightarrow 2
  4. DLyman, ∞→1\infty \rightarrow 1

Correct answer

A. Paschen, ∞→3\infty \rightarrow 3

Step-by-step solution

1λ=R\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)n1=3,n2=∞n_{1}=3, n_{2}=\infty 1λ=R(19)⇒λ=9R=9105=9×10−5 cm=900nm\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{9}\right) \Rightarrow \lambda=\frac{9}{\mathrm{R}}=\frac{9}{10^{5}}=9 \times 10^{-5} \mathrm{~cm}=900 \mathrm{nm}
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