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 $\left.\mathrm{R}_{\mathrm{H}}=10^5 \mathrm{~cm}^{-1}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$

<p>To determine which spectral line of the hydrogen atom suits the wavelength for heat treatment of muscular pain (900 nm), we begin by applying the Rydberg equation:</p> <p>$ \frac{1}{\lambda} = \mathrm{R}_{\mathrm{H}} \mathrm{Z}^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $</p> <p>Given:</p> <p><p>$\lambda = 900 \text{ nm} = 9 \times 10^{-5} \text{ cm}$</p></p> <p><p>$\mathrm{R}_{\mathrm{H}} = 10^5 \text{ cm}^{-1}$</p></p> <p><p>Hydrogen atom (Z = 1)</p></p> <p>We proceed by solving the equation:</p> <p>$ \frac{1}{\lambda \times \mathrm{R}_{\mathrm{H}}} = \frac{1}{n_1^2} - \frac{1}{n_2^2} $</p> <p>Substituting the known values:</p> <p>$ \frac{1}{9 \times 10^{-5} \text{ cm} \times 10^5 \text{ cm}^{-1}} = \frac{1}{n_1^2} - \frac{1}{n_2^2} $</p> <p>This simplifies to:</p> <p>$ \frac{1}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2} $</p> <p>This condition is satisfied when $ n_1 = 3 $ and $ n_2 = \infty $.</p> <p>Therefore, the appropriate spectral transition is from $ n_2 = \infty $ to $ n_1 = 3 $, which corresponds to the Paschen series transition $ \infty \rightarrow 3 $.</p>