Question

Bond energy of ${Br}_{2}is194kJ/mol$. Find the minimum wave no. of photon required to break this bond ?

Answer 1

Bond dissociation energy= $194\times {10}^{3}J/mol$

Therefore, the energy required to break the br-br bond=$\frac{194\times {10}^3}{6.022\times {10}^{23}}$

Now, E=hv

$v=\frac{E}{h}=\frac{194\times {10}^{3}J}{6.022\times {10}^{23}\times 6.022\times {10}^{-34}j-s}$

= $\mathrm{4..8743}\times {10}^{14}{s}^{-1}$