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Iodine molecules dissociate into atoms after absorbing light of wavelength $4500 Å$ . If one quantum of radiation is absorbed by each molecule, what will be the approximate kinetic energy of one iodine atom? [Given B.E. of $I_{2} = 240 kJ / mol], [= 6 \times 10^{23}]$ ?

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$4.3 J$

$8.6 \times 10^{- 36} J$

$2.17 \times 10^{- 20} J$

$433 kJ$

answer is B.

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Detailed Solution

The energy of incident;

$E_{incident} = \frac{12400}{4500} \times 96.48 kJ / mol = 266 kJ / mol$

KE of 2 mole iodine atoms = 266 - 240 = 26 kJ

$∴ K E$ of 1 iodine atom $= \frac{26 \times 10^{3}}{2 \times 6 \times 10^{23}} = 2.17 \times 10^{- 20} J$ .

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