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Structure of Atom question

2019 · 11 Jan · Shift 1 · Q7
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Structure of Atom question

2019 · 11 Jan · Shift 1 · Q7

JEE MainChemistryStructure of AtomMCQ+4 / −1
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]
  1. A
    Balmer, ∞→\infty \to∞→ 2
  2. B
    Paschen, 5 →\to→ 3
  3. C
    Paschen, ∞→\infty \to∞→ 3
  4. D
    Lyman, ∞→\infty \to∞→ 1
View written solutionFree

Correct answer: C

  1. Use the Rydberg formula for hydrogen spectral lines:
νˉ=1λ=RH(1n12−1n22),n2>n1\bar\nu = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1νˉ=λ1​=RH​(n12​1​−n22​1​),n2​>n1​

We need a wavelength near 900 nm900\,\text{nm}900nm.

Convert this to cm:

900 nm=900×10−9 m=9×10−7 m900\,\text{nm} = 900\times 10^{-9}\,\text{m} = 9\times 10^{-7}\,\text{m}900nm=900×10−9m=9×10−7m

Since 1 m=100 cm1\,\text{m}=100\,\text{cm}1m=100cm,

9×10−7 m=9×10−5 cm9\times 10^{-7}\,\text{m} = 9\times 10^{-5}\,\text{cm}9×10−7m=9×10−5cm

So,

νˉ=1λ=19×10−5≈1.11×104 cm−1\bar\nu = \frac{1}{\lambda} = \frac{1}{9\times 10^{-5}} \approx 1.11\times 10^4\,\text{cm}^{-1}νˉ=λ1​=9×10−51​≈1.11×104cm−1
  1. Check each option using RH=105 cm−1R_H = 10^5\,\text{cm}^{-1}RH​=105cm−1.

Option A: Balmer, ∞→2\infty \to 2∞→2

νˉ=105(122−1∞2)=105(14)=2.5×104 cm−1\bar\nu = 10^5\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=10^5\left(\frac14\right)=2.5\times 10^4\,\text{cm}^{-1}νˉ=105(221​−∞21​)=105(41​)=2.5×104cm−1 λ=12.5×104=4×10−5 cm=400 nm\lambda = \frac{1}{2.5\times 10^4}=4\times 10^{-5}\,\text{cm}=400\,\text{nm}λ=2.5×1041​=4×10−5cm=400nm

Not close to 900 nm900\,\text{nm}900nm.

Option B: Paschen, 5→35 \to 35→3

νˉ=105(132−152)=105(19−125)\bar\nu = 10^5\left(\frac{1}{3^2}-\frac{1}{5^2}\right) =10^5\left(\frac19-\frac1{25}\right)νˉ=105(321​−521​)=105(91​−251​) 19−125=25−9225=16225\frac19-\frac1{25} = \frac{25-9}{225}=\frac{16}{225}91​−251​=22525−9​=22516​

So,

νˉ=105⋅16225≈7.11×103 cm−1\bar\nu = 10^5\cdot \frac{16}{225} \approx 7.11\times 10^3\,\text{cm}^{-1}νˉ=105⋅22516​≈7.11×103cm−1 λ≈17.11×103=1.406×10−4 cm=1406 nm\lambda \approx \frac{1}{7.11\times 10^3}=1.406\times 10^{-4}\,\text{cm}=1406\,\text{nm}λ≈7.11×1031​=1.406×10−4cm=1406nm

Not close to 900 nm900\,\text{nm}900nm.

Option C: Paschen, ∞→3\infty \to 3∞→3

νˉ=105(132−0)=105(19)=1.11×104 cm−1\bar\nu = 10^5\left(\frac{1}{3^2}-0\right)=10^5\left(\frac19\right)=1.11\times 10^4\,\text{cm}^{-1}νˉ=105(321​−0)=105(91​)=1.11×104cm−1 λ=11.11×104=9×10−5 cm=900 nm\lambda = \frac{1}{1.11\times 10^4} = 9\times 10^{-5}\,\text{cm}=900\,\text{nm}λ=1.11×1041​=9×10−5cm=900nm

This matches exactly.

Option D: Lyman, ∞→1\infty \to 1∞→1

νˉ=105(1−0)=105 cm−1\bar\nu = 10^5\left(1-0\right)=10^5\,\text{cm}^{-1}νˉ=105(1−0)=105cm−1 λ=1105=10−5 cm=100 nm\lambda = \frac{1}{10^5}=10^{-5}\,\text{cm}=100\,\text{nm}λ=1051​=10−5cm=100nm

Not close to 900 nm900\,\text{nm}900nm.

  1. Conclusion

The spectral line corresponding to wavelength ≈900 nm\approx 900\,\text{nm}≈900nm is:

Paschen, ∞→3\boxed{\text{Paschen, } \infty \to 3}Paschen, ∞→3​

So the correct option is C.

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