Question

A photon having wavelength equal to 500 nm is used for photolysis of a molecule X2 having bond energy of 144 kJ/mol. The % of the energy of photon which is converted into the kinetic energy of X atoms is X. The value of is (to the nearest integer) [hc = 1240 eV nm, 1 eV/molecule = 96 kJ/mol] 10 =​

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Answer 1

A photon having wavelength equal to 500 nm is used for photolysis of a molecule X₂ having bond energy of 144 kJ/mol.

We have to find % of the energy of photon is converted into the kinetic energy of X atoms.

incident photon strikes on the molecule X₂ , after that X₂ is broken into its constituent elements gaining bond energy from photon. and then rest of energy of photon provides them to get kinetic energy of its atoms.

∴ kinetic energy = energy of incident photon - bond energy of X₂

we know, energy of incident photon is expressed as E = $\bf\frac{hc}{\lambda}$

here, hc = 1240 ev nm and λ = 500 nm

so the energy of incident photon, E = $\frac{1240}{500}$ = 2.48 eV/molecule

here , 1 eV/molecule = 96 kJ/mol

∴ E = 2.48 × 96 = 238.08 kJ/mol

Bond energy of X₂ = 144 kJ/mol

∴ kinetic energy of X atoms = 238.08 - 144 = 94.08 kJ/mol

now the percentage of the energy of photon is converted into kinetic enery = $\frac{\text{kinetic energy of X atoms}}{\text{energy of incident photon}}\times100$

= $\frac{94.08}{238.08}\times100$

≈ 40 %

Therefore 40 % of the energy of photon which is converted inot the kinetic energy of X atoms.