A photon of 4000 oA is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of I2 molecule is 246.5 kJ/mol is:
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 one photon =124004000=3.1ev=3.11×6×10−19Joules Energy supplied by one mole photon in KJ/mole =3.11×6×10−19×6×1023×10−3=297KJ/mole % of energy converted to K.E.=297−246.5297=17