Question

Photon having wavelength $310 nm$ is used to break the bond of $A _{2}$ molecule having bond energy $288 kJ / mol$ then the percentage of energy of photon converted to $KE$ is $(1 eV =96 kJ / mol )$

• A. 25
• B. 50
• C. 75
• D. 80

Answer 1

Answer: (A)

$E=h v$

$E =\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{310 \times 10^{-9}} J$

$E =0.0641 \times 10^{-26} \times 10^{9} J$

$E=0.0641 \times 10^{-17} J$

This is the energy of one photon required to break one $A _{2}$ molecule.

To break 1 mole of $A_{2}$, Energy of photons required =

$=6.02 \times 10^{23} \times 0.0641 \times 10^{-17}=0.386 \times 10^{6}=386 kJ / mole$

$K . E =386 kJ / mole -288 kJ / mole =98 kJ / mole$

Percentage of total energy that got converted into $KE =$

$\frac{98}{386} \times 100=25 \%$