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Q.

A photon with a wavelength of 4000A° 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

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a

12%

b

25%

c

17%

d

8%

answer is C.

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Detailed Solution

Energy of one photon = 124004000=3.1eV
Energy supplied by one mole photon in KJ/mole =3.1×1.6×10−19×1023×10−3=297kJmol−1
∴% of energy converted to
K.E=297−246.5297=17%

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