Photon having wavelength 310nm is used to break the bond of A2 molecule having bond energy 288 kj mol−1then % energy of photon converted to the KE is :[hc=12400eVA,1eV=96kJ/mol]

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Photon having wavelength 310nm is used to break the bond of $A_{2}$ molecule having bond energy $288 k j {mol}^{- 1}$ then % energy of photon converted to the KE is : $[h c = 12400 eVA, 1 eV = 96 kJ / mol]$

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Detailed Solution

Incident energy;

$E_{incident} = \frac{1240}{310} = 4 eV; E_{bond} = \frac{288}{96} = 3 eV$

$∴ K E = 1 eV$

$∴ % converted = 25 %$

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