Question

Bond dissociation energy of a bond (A - B) is $165kJ{}^{-1}$. Calculate the minimum frequency of photon required to break this bond.

Answer 1

Bond dissociation energy (E) = $165\times {10}^{3}J$

Energy of photon $=h\nu$

$E=h\nu$

$165\times {10}^3=6.622\times {10}^{-34}\times \nu$

$\nu =\frac{165\times {10}^3}{6.626\times {10}^{-34}}$

$\nu =24.9\times {10}^{37}$ Hz

$\nu =2.49\times {10}^{38}$ HZ

Therefore the minimum frequency of photon required to break the bond is $2.49\times {10}^{38}$