4000∘A photon is used to break iodine molecule, then the % of energy converted to K.E of Iodine atoms, if the bond dissociation energy of I2 molecule is 246.5kJmol−1
A
8%
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B
12%
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C
17%
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D
25%
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Solution
The correct option is C 17% Energy of photo = hcλ=6.625×10−34×3×108×6.023×10284000×10−10=299.26KJ/mole % of energy converted into K.E = 299−246.5299×100=17%approx