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Chemistry Level 4

Iodine molecule dissociates into atoms after absorbing light of 450 $\text{nm}$ . If one quantum of radiation is absorbed by each molecule, calculate the kinetic energy of each iodine atom.

If it can be written as $a \times 10^b \text{ J}$ where $0 \leq a <10$ , find the value of $\lfloor a \rfloor |b|$ .

Details and Assumptions

$c= 3.0 \times 10^{8} \text{ m/s} \\ h = 6.63 \times 10^{-34} \text{ Js} \\ 1 \text{ mol}= 6.02 \times 10^{23} \text{ entities} \\ \text{Bond energy of }\ce{I_2}= 240 \text{ kJ mol}^{-1}$

The answer is 40.

2 solutions

Akshat Sharda
Jun 14, 2016

Bond energy of $\text{I}_{2}$ is approximately $2.487\text{ eV}$ per molecule.

The energy of one photon of $450 \text{ nm}$ wavelength is approximately $2.755\text{ eV}$ .

Therefore, the molecule takes $2.487\text{ eV}$ to disassociate into 2 atoms and then each atom possess kinetic energy equivalent to, $\frac{2.755-2.287}{2} \text{ eV}=0.134\text{ eV}\sim 2.144×10^{-20} \text{J} \\ \Rightarrow \lfloor 2.144\rfloor |-20|=2×20=\boxed{40}$

Tom Van Lier
Jun 30, 2016

The energy associated with one quantum equals $E = h \nu = \dfrac{hc}{\lambda} = 4.42 \times 10^{-19} J$ .

The energy needed to break down one $I_{2}-$ molecule is equal to $E´= \dfrac{240000 }{6.02 \times 10^{23}} J = 3.98671096 \times 10^{-19} J$ .

So each molecule gets $E - E' = 4.33289 \times 10^{-20} J$ of kinetic energy, which means each atom gets half.

So $\lfloor a \rfloor = 2 \wedge |b| = 20$ and thus the answer is 40.

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The answer is 40.

2 solutions

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