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Atoms and Nuclei question

2010 · Shift 2 · Q55
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Atoms and Nuclei question

2010 · Shift 2 · Q55

JEE AdvancedPhysicsAtoms and NucleiMCQ+3 / −1
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu =53×10−27= {5 \over 3} \times {10^{ - 27}}=35​×10−27 kg, is close to :
  1. A
    2.4 ×\times× 10 −-− 10 m
  2. B
    1.9 ×\times× 10 −-− 10 m
  3. C
    1.3 ×\times× 10 −-− 10 m
  4. D
    4.4 ×\times× 10 −-− 11 m
View written solutionFree

Correct answer: C

Step-by-step Solution:

  1. Analyze the Question: The problem asks for the interatomic distance in a carbon monoxide (CO) molecule. It provides an introductory paragraph about applying Bohr's quantization condition to molecular rotation. However, it gives no numerical data, such as rotational energy levels or transition frequencies, that would allow for a direct calculation of the bond length from this principle. This suggests that either the question is knowledge-based, requiring you to know the approximate size of a CO molecule, or it is part of a larger problem set with missing data. We will solve it by comparing the known value of the CO bond length with the given options.

  2. Recall or Estimate the Bond Length: Atomic bond lengths are typically on the order of Angstroms (1A˚=10−101 \AA = 10^{-10}1A˚=10−10 m). For a CO molecule, the bond is a triple bond, which is short and strong. The experimentally determined bond length of a CO molecule is approximately 1.128A˚1.128 \AA1.128A˚, or 1.128×10−101.128 \times 10^{-10}1.128×10−10 m.

  3. Calculate the Reduced Mass (for context, not calculation of r): Although we cannot calculate the distance r without more data, we can calculate the reduced mass μ\muμ of the CO molecule to confirm the parameters are standard.

    • Mass of Carbon, mC=12m_C = 12mC​=12 amu
    • Mass of Oxygen, mO=16m_O = 16mO​=16 amu
    • Given conversion: 1 amu=53×10−271 \text{ amu} = \frac{5}{3} \times 10^{-27}1 amu=35​×10−27 kg

    The formula for the reduced mass is: μ=mCmOmC+mO\mu = \frac{m_C m_O}{m_C + m_O}μ=mC​+mO​mC​mO​​ Substituting the values in amu: μ=12×1612+16 amu=19228 amu=487 amu\mu = \frac{12 \times 16}{12 + 16} \text{ amu} = \frac{192}{28} \text{ amu} = \frac{48}{7} \text{ amu}μ=12+1612×16​ amu=28192​ amu=748​ amu Converting to kg: μ=487×(53×10−27 kg)=16×57×10−27 kg=807×10−27 kg≈11.43×10−27 kg\mu = \frac{48}{7} \times \left( \frac{5}{3} \times 10^{-27} \text{ kg} \right) = \frac{16 \times 5}{7} \times 10^{-27} \text{ kg} = \frac{80}{7} \times 10^{-27} \text{ kg} \approx 11.43 \times 10^{-27} \text{ kg}μ=748​×(35​×10−27 kg)=716×5​×10−27 kg=780​×10−27 kg≈11.43×10−27 kg This is the correct reduced mass for a ¹²C¹⁶O molecule.

  4. Compare the Known Bond Length with the Options:

    • Known value: ractual≈1.13×10−10r_{actual} \approx 1.13 \times 10^{-10}ractual​≈1.13×10−10 m
    • Option A: 2.4×10−102.4 \times 10^{-10}2.4×10−10 m
    • Option B: 1.9×10−101.9 \times 10^{-10}1.9×10−10 m
    • Option C: 1.3×10−101.3 \times 10^{-10}1.3×10−10 m
    • Option D: 4.4×10−114.4 \times 10^{-11}4.4×10−11 m = 0.44×10−100.44 \times 10^{-10}0.44×10−10 m

  5. Select the Closest Option: We need to find which option is closest to the actual value of 1.13×10−101.13 \times 10^{-10}1.13×10−10 m.

    • ∣1.3×10−10−1.13×10−10∣=0.17×10−10|1.3 \times 10^{-10} - 1.13 \times 10^{-10}| = 0.17 \times 10^{-10}∣1.3×10−10−1.13×10−10∣=0.17×10−10 m
    • ∣1.9×10−10−1.13×10−10∣=0.77×10−10|1.9 \times 10^{-10} - 1.13 \times 10^{-10}| = 0.77 \times 10^{-10}∣1.9×10−10−1.13×10−10∣=0.77×10−10 m
    • ∣2.4×10−10−1.13×10−10∣=1.27×10−10|2.4 \times 10^{-10} - 1.13 \times 10^{-10}| = 1.27 \times 10^{-10}∣2.4×10−10−1.13×10−10∣=1.27×10−10 m
    • ∣0.44×10−10−1.13×10−10∣=0.69×10−10|0.44 \times 10^{-10} - 1.13 \times 10^{-10}| = 0.69 \times 10^{-10}∣0.44×10−10−1.13×10−10∣=0.69×10−10 m

    Option C is numerically the closest to the true value. While there is a discrepancy of about 15%, the other options are significantly farther off. In the context of a multiple-choice question, this is the most plausible answer.

Conclusion:

The most reasonable answer is the one closest to the known experimental value for the bond length of a CO molecule (1.13×10−101.13 \times 10^{-10}1.13×10−10 m). Option C, 1.3×10−101.3 \times 10^{-10}1.3×10−10 m, is the best fit among the choices provided.

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