A UV photon of wavelength 215 nm is used to photolyse O2, whose O=O bond energy is 498 kJ mol^-1. Take h = 6.63x10^-34 J s, c = 3x10^8 m s^-1, N_A = 6.02x10^23 mol^-1. Calculate the excess energy, per mole of photons, left after breaking the bond (in kJ mol^-1, nearest integer).

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JEE Advanced levelNumerical answerDifficulty 3 of 3ev110

A UV photon of wavelength $215 nm$ is used to photolyse $O X_{2}$ , whose O=O bond energy is $498 kJ mo l^{- 1}$ . Take $h = 6.63 \times 1 0^{- 34} J s$ , $c = 3 \times 1 0^{8} m s^{- 1}$ , $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ . Calculate the excess energy, per mole of photons, left after breaking the bond (in $kJ mo l^{- 1}$ , nearest integer).

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Answer: 59 kJ/mol

Energy per mole of photons =

h c N_{A} / λ

= (6.63e-34 x 3e8 x 6.02e23)/(215e-9) J = 556.9 kJ/mol. Excess = 556.92 - 498 = 58.92 kJ/mol, which is 59 kJ/mol to the nearest integer. This excess appears as kinetic energy of the O atoms.

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A UV photon of wavelength 215 nm is used to photolyse OX2​, whose O=O bond energy is 498 kJ mol−1. Take h=6.63×10−34 J s, c=3×108 m s−1, NA​=6.02×1023 mol−1. Calculate the excess energy, per mole of photons, left after breaking the bond (in kJ mol−1, nearest integer).

Answer: 59 kJ/mol

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