Question

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

• A. 25
• B. 75
• C. 50
• D. 80

Answer 1

The correct option is A 25

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

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

Energy of photon converted to kinetic energy KE $=(384-288)$ $kJ/mol$
$=96$ $kJ/mol$

$\therefore$ % energy of photon converted into KE $=\frac{96}{383}\times 100=25\%$