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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 = $\frac{12400}{4000} = 3.1 eV$
Energy supplied by one mole photon in KJ/mole $= 3.1 \times 1.6 \times 10^{- 19} \times 10^{23} \times 10^{- 3} = 297 kJ {mol}^{- 1}$
$∴$ % of energy converted to
$K . E = \frac{297 - 246.5}{297} = 17 %$

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