Question

Photon having wavelength $310\text{nm}$ is used to break the bond of $A_2$ molecule having bond energy $288{\text{kJ mol}}^{–1}$, then percentage of energy of photon converted to the kinetic energy is:
$[hc=12400{\text{evA}}^{\circ },1\text{ev}=96\text{kJ/mol}]$

• A. $25\%$
• B. $50\%$
• C. $75\%$
• D. $80\%$

Answer 1

The correct option is A $25\%$

We know, energy of one photon $=\frac{hc}{\lambda }$
Where,
$h=\text{Planck's constant}$
$\lambda =\text{Wavelength}$
Given, $hc=12400{\text{evA}}^{\circ }$
$\therefore \text{Energy of one photon}=\frac{12400}{3100}=4\text{ev}$
$\text{Hence energy of photon}=(4\times 96)\text{kJ/mol}=384\text{kJ/mol}$
$\therefore$ percentage of energy converted to kinetic energy
$=\frac{384-288}{384}=\frac{96}{384}\times 100=25\%$