Question

Photon having wavelength $310\text{nm}$ is used to break the bond of $A_2$ molecule having bond energy $288{\text{kJmol}}^{-1}$, the $\%$ energy of photon converted to kinetic energy is:
$(hc=12400eV\stackrel{\circ }{A}$ , $1eV=96\text{kJ/mol})$

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

Answer 1

The correct option is A $25\%$

Given :
${\lambda }_{photon}=310\text{nm}$
$\text{Bond energy}=288\text{kJ/mol}$
$hc=12400eV\stackrel{\circ }{A}$
$=1.19\times {10}^{-4}{\text{kJ m mol}}^{-1}$
$1eV=96$ $kJ/mol$

Energy of 1 photon $=\frac{hc}{{\lambda }_{photon}}$
$=\frac{1.19\times {10}^{-4}}{310\times {10}^{-9}}$ $=384$ $\text{kJ/mol}$

Energy of photon converted to kinetic energy KE $=(384-288)\text{kJ/mol}=96\text{kJ/mol}$
$\therefore \%$ energy of photon converted into KE $=\frac{96}{384}\times 100=25\%$