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

(B) $\frac{\mathrm{N}_{\mathrm{A}}\mathrm{hc}}{250(4\mathrm{y}-6\mathrm{x})}$

(A) $\frac{\mathrm{hc}}{1000}\left(\frac{\mathrm{y}-6\mathrm{x}}{4}\right)^{-1}$

(C) $\frac{\mathrm{N}_{\mathrm{A}}\mathrm{hc}}{250(\mathrm{y}-6\mathrm{x})}$

(D) $N_{A}hc\left(y-\frac{6x}{4}\right)^{-1}$

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Chemical Thermodynamics

The heat of atomisation of methane and ethane are 'x'

kJ mol^{- 1}

and 'y'

kJ mol^{- 1}

respectively. The longest wavelength (

λ

) of light capable of breaking the

C - C

bond can be expressed in SI unit as:

Answer: (B)

Heat of atomisation is the energy to break every bond in one mole. For methane

C H_{4} \to C + 4 H

, so

x = 4 ε_{C - H}

(kJ), giving

ε_{C - H} = \frac{x}{4} kJ = 250 x J

. For ethane

C_{2} H_{6} \to 2 C + 6 H

, so

y = ε_{C - C} + 6 ε_{C - H}

(kJ), hence

ε_{C - C} = 1000 y - 6 (250 x) = 250 (4 y - 6 x) J mol^{- 1}

. The longest wavelength able to break one C-C bond satisfies

\frac{h c}{λ} = \frac{ε _{C - C}}{N _{A}}

, so

λ = \frac{N _{A} h c}{ε _{C - C}} = \frac{N _{A} h c}{250 ( 4 y - 6 x )}

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option