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Question

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=3.11×6×1019Joules
Energy supplied by one mole photon in KJ/mole =3.11×6×1019×6×1023×103=297KJ/mole
% of energy converted to K.E.=297246.5297=17

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