Question

Bond energy of $F_2$ is $150$ $kJ$ ${mol}^{-1}$. Calculate the minimum frequency of photon to break this bond.

Answer 1

Bond energy of $1$ molecule of $F_2=\frac{B.E.\left(BondEnergy\right)}{N_A\left(Avagadronumber\right)}=\frac{150\times 1000}{6.023\times {10}^{23}}=24.9\times {10}^{-20}J/mol=1.55eVBE=hv24.9\times {10}^{-20}=6.626\times {10}^{-34}\times v\Rightarrow v=3.759\times {10}^{14}Hz$