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QJEEM25Ch17ef: JEE Main 2025 Chemistry - Class 11 JEEM25Ch17ef Solution: Heat treatment of muscular pain involves radiation of…

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm . Which spectral line of H atom is suitable for this?

Given : Rydberg constant RH=105 cm−1, h=6.6×10−34 J s,c=3×108 m/s)\left.\mathrm{R}_{\mathrm{H}}=10^5 \mathrm{~cm}^{-1}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)

Multiple Choice Options

  • A Lyman series, ∞→1\infty \rightarrow 1
  • B Balmer series, ∞→2\infty \rightarrow 2
  • C Paschen series, 5→35 \rightarrow 3
  • D Paschen series, ∞→3\infty \rightarrow 3

Step-by-Step Verified Solution

λ=900 nmamp; H-atom (Z=1)=9×10−5 cmRH=105 cm−1amp;\begin{array}{ll}\lambda=900 \mathrm{~nm} & \text { H-atom }(\mathrm{Z}=1) \\=9 \times 10^{-5} \mathrm{~cm} \\\mathrm{R}_{\mathrm{H}}=10^5 \mathrm{~cm}^{-1} &\end{array}

amp; Ryderg eq. =1λ=RHZ2×(1n12−1n22)amp;⇒1λ×RH=1n12−1n22amp;⇒19×10−5 cm×105 cm−1=(1n12−1n22)amp;⇒1n12−1n22=19\begin{aligned}& \text { Ryderg eq. }=\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}} \mathrm{Z}^2 \times\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \\& \Rightarrow \frac{1}{\lambda \times \mathrm{R}_{\mathrm{H}}}=\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2} \\& \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) \\& \Rightarrow \frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}=\frac{1}{9}\end{aligned}

It is possible when n1=3,n2=∞n_1=3, n_2=\infty

Possible series: ∞→3\infty \rightarrow 3

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