Question 3b535

Apr 18, 2017

2.16×10^-20J

Explanation:

Energy given to each ${\text{I}}_{2}$ molecule through one quantum of radiationE=(hc)/λ
where $h$ is Planck's Constant, $c$ is velocity of light in vacuum and $\lambda$ wave length of light quantum.

Inserting various values we get
E=(6.626×10^-34×3×10^8)/(4500×10^-10)
⇒E=4.417×10^-19J

We know that one mole of ${\text{I}}_{2}$ molecule$=$Avogadro's number.
$\therefore$Energy used for breaking up of each ${\text{I}}_{2}$ molecule=(2.4×10^3)/(6.023×10^23)
=3.984×10^-19J

Balance energy is used in imparting kinetic energy to two Iodine atomsE_b=(4.417-3.984)×10^-19J
Kinetic energy of each Iodine atom E_b/2=(4.417−3.984)/2×10^-19
=2.16×10^-20J#