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QJEEM19Ch3cac: JEE Main 2019 Chemistry - Class 11 JEEM19Ch3cac 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 purpose?

[RH = 1 ×\times 105 cm–1, h = 6.6 ×\times 10–34 Js, c = 3 ×\times 108 ms–1]

Multiple Choice Options

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

Step-by-Step Verified Solution

Given, RH = 1 ×\times 105 cm–1

⇒\Rightarrow 1RH{1 \over {{R_H}}} = 10-5 cm

⇒\Rightarrow 1RH{1 \over {{R_H}}} = 10-7 cm ×\times 100

⇒\Rightarrow 1RH{1 \over {{R_H}}} = 100 nm

We know,

1λ=ν=RH×Z2(1nL2−1nH2){1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)

⇒\Rightarrow λ\lambda = 1RH×(1)2×1(1nL2−1nH2){1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

[For H atom Z = 1]

⇒\Rightarrow λ=1RH×1(1nL2−1nH2)\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

⇒\Rightarrow λ\lambda = 100(1nL2−1nH2){{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}

Given, λ\lambda = 900 nm

∴\therefore 100(1nL2−1nH2){{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}} = 900

⇒\Rightarrow (1nL2−1nH2)=19{\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}

By checking each options you can see

when nL = 3 and nH = ∞\infty then

(1nL2−1nH2)=19{\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}

∴\therefore Option C is correct.

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