Question

Photon having wavelength $310nm$ is used to break the bond of $A_2$ molecule having bond energy $288kJ/mol$ then the percentage of energy of photon converted to $KE$ is $(1eV=96kJ/mol)$

• A. 25
• B. 50
• C. 75
• D. 80

Answer 1

Answer: (A)

$E=hv$

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

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

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

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 =

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

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

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

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