Q6 / 75 · JEE Main 2026 (24 January Shift 2) · Chemistry
The heat of atomisation of methane and ethane are ' x ' $\mathrm{kJ} \mathrm{mol}^{-1}$ and ' y ' $\mathrm{kJ} \mathrm{mol}^{-1}$ respectively. The longest wavelength $(\lambda)$ of light capable of breaking the $\mathrm{C}-\mathrm{C}$ bond can be expressed in SI unit as:
- Correct answerA$\frac{\mathrm{N}_{\mathrm{A}} \mathrm{hc}}{250(4 y-6 x)}$
- B$\frac{\mathrm{hc}}{1000}\left(\frac{y-6 x}{4}\right)^{-1}$
- C$\frac{\mathrm{N}_{\mathrm{A}} \mathrm{hc}}{250(y-6 x)}$
- D$\mathrm{N}_{\text {A }} \mathrm{hc}\left(y-\frac{6 x}{4}\right)^{-1}$
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 $E = \frac{hc}{\lambda}$ and $\lambda = \frac{hc}{E}$:
$\lambda = \frac{N_A hc}{250(4y - 6x)}$ (in SI units)
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 $E = \frac{hc}{\lambda}$ and $\lambda = \frac{hc}{E}$:
$\lambda = \frac{N_A hc}{250(4y - 6x)}$ (in SI units)
Hence, the correct answer is option: 'A'.