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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2026 · 24 January · Evening Shift · Chemistry · Thermodynamics (C)

The heat of atomisation of methane and ethane are ' x ' kJ mol^ -1...

The heat of atomisation of methane and ethane are ' x ' kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} and ' y ' kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} respectively. The longest wavelength (λ)(\lambda) of light capable of breaking the C−C\mathrm{C}-\mathrm{C} bond can be expressed in SI unit as:

Options

  1. ANAhc250(4y−6x)\frac{\mathrm{N}_{\mathrm{A}} \mathrm{hc}}{250(4 y-6 x)}
  2. B\frac{\mathrm{hc}}{1000}(y−6x4)−1\left(\frac{y-6 x}{4}\right)^{-1}
  3. CNAhc250(y−6x)\frac{\mathrm{N}_{\mathrm{A}} \mathrm{hc}}{250(y-6 x)}
  4. D\mathrm{N}_{\text{A }} \mathrm{hc}(y−6x4)−1\left(y-\frac{6 x}{4}\right)^{-1}

Correct answer

A. NAhc250(4y−6x)\frac{\mathrm{N}_{\mathrm{A}} \mathrm{hc}}{250(4 y-6 x)}

Step-by-step solution

Heat of atomization: CH₄ has x kJ/mol (4 C-H bonds), C₂H₆ has y kJ/mol (1 C-C + 6 C-H bonds)
Let C-H bond energy = a, C-C bond energy = b
Then: x = 4a and y = b + 6a
Solving: a = x/4, so b = y - 6x/4 = (4y - 6x)/4 kJ/mol
Converting to J/mol: b = (4y - 6x) × 250 J/mol
Using E=hcλE = \frac{hc}{\lambda} and λ=hcE\lambda = \frac{hc}{E}:
λ=NAhc250(4y−6x)\lambda = \frac{N_A hc}{250(4y - 6x)} (in SI units)
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