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Chemistry · Structure of Atom

JEE Main 2025 — 23 January, Morning Shift — Question 27

Heat treatment of muscular pain involves radiation of wavelength about 900 nm900\,\mathrm{nm}. Which spectral line of hydrogen atom is suitable?

(Given: RH=105 cm−1R_H = 10^5\,\mathrm{cm^{-1}})

  1. Option A:

    Paschen series, ∞→3\infty \rightarrow 3

    Correct
  2. Option B:

    Lyman series, ∞→1\infty \rightarrow 1

  3. Option C:

    Balmer series, ∞→2\infty \rightarrow 2

  4. Option D:

    Paschen series, 5→35 \rightarrow 3

Answer: A

Step-by-step solution

λ=900 nm\lambda=900 \mathrm{~nm} H-atom ( Z=1\mathrm{Z}=1 ) =9×10−5 cm=9 \times 10^{-5} \mathrm{~cm}

RH=105 cm−1\mathrm{R}_{\mathrm{H}}=10^{5} \mathrm{~cm}^{-1}

Ryderg eq. =1λ=RHZ2×(1n12−1n22)=\frac{1}{\lambda}=R_{H} Z^{2} \times\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)

⇒1λ×RH=1n12−1n22\Rightarrow \frac{1}{\lambda \times \mathrm{R}_{\mathrm{H}}}=\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}

⇒19×10−5 cm×105 cm−1=(1n12−1n22)\Rightarrow \frac{1}{9 \times 10^{-5} \mathrm{~cm} \times 10^{5} \mathrm{~cm}^{-1}}=\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)

⇒1n12−1n22=19\Rightarrow \frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}=\frac{1}{9}

It is possible when n1=3,n2=∞\mathrm{n}_{1}=3, \mathrm{n}_{2}=\infty

Possible series : ∞→3\infty \rightarrow 3

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Chemistry
Chapter
Structure of Atom
Topic
Analysis of Spectra of H-like species