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Structure of Atom question

2010 · Shift 0 · Q26
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Structure of Atom question

2010 · Shift 0 · Q26

JEE MainChemistryStructure of AtomMCQ+4 / −1
The energy required to break one mole of Cl–Cl bonds in Cl2Cl_2Cl2​ is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 x 108 ms–1 and NA = 6.02 x 1023 mol–1)
  1. A
    594 nm
  2. B
    640 nm
  3. C
    700 nm
  4. D
    494 nm
View written solutionFree

Correct answer: D

  1. Energy required for one bond

Given bond dissociation energy for one mole of Cl–Cl\text{Cl–Cl}Cl–Cl bonds:

Emol=242 kJ mol−1=242×103 J mol−1E_{\text{mol}} = 242\ \text{kJ mol}^{-1} = 242 \times 10^3\ \text{J mol}^{-1}Emol​=242 kJ mol−1=242×103 J mol−1

Energy for breaking one Cl–Cl\text{Cl–Cl}Cl–Cl bond is:

E=242×1036.02×1023 JE = \frac{242 \times 10^3}{6.02 \times 10^{23}}\ \text{J}E=6.02×1023242×103​ J

E≈4.02×10−19 JE \approx 4.02 \times 10^{-19}\ \text{J}E≈4.02×10−19 J

  1. Use photon energy relation

For the longest wavelength capable of breaking the bond, photon energy should be exactly equal to bond energy:

E=hcλE = \frac{hc}{\lambda}E=λhc​

So,

λ=hcE\lambda = \frac{hc}{E}λ=Ehc​

Given:

  • h=6.626×10−34 J sh = 6.626 \times 10^{-34}\ \text{J s}h=6.626×10−34 J s
  • c=3×108 m s−1c = 3 \times 10^8\ \text{m s}^{-1}c=3×108 m s−1

Thus,

λ=(6.626×10−34)(3×108)4.02×10−19\lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4.02 \times 10^{-19}}λ=4.02×10−19(6.626×10−34)(3×108)​

λ≈4.94×10−7 m\lambda \approx 4.94 \times 10^{-7}\ \text{m}λ≈4.94×10−7 m

  1. Convert into nm

1 nm=10−9 m1\ \text{nm} = 10^{-9}\ \text{m}1 nm=10−9 m

λ=4.94×10−7 m=494 nm\lambda = 4.94 \times 10^{-7}\ \text{m} = 494\ \text{nm}λ=4.94×10−7 m=494 nm

  1. Match with options

The correct option is:

D: 494 nm\boxed{\text{D: }494\ \text{nm}}D: 494 nm​

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