Question

bond energy of h2 is 436KJ/mol . determine the longest wavelength of photons which when to incident upon a sample of h2 will bring about photochemical decomposition followed by excitation to fifth bohrs orbit .

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Bond}\mathrm{energy}\mathrm{per}\mathrm{molecule}=\frac{436\times {10}^3\mathrm{J}/\mathrm{mol}}{6.023\times {10}^{23}}=7.24\times {10}^{-19}\mathrm{J} \\ \mathrm{Excitation}\mathrm{energy}/\mathrm{H}-\mathrm{atom}=2.18\times {10}^{-18}(\frac{1}{1^2}-\frac{1}{5^2})\mathrm{J}=2.18\times {10}^{-18}\times \frac{24}{25}\mathrm{J}=2.09\times {10}^{-18}\mathrm{J} \\ \mathrm{In}\mathrm{the}\mathrm{process}\mathrm{of}\mathrm{decoposition},\mathrm{one}\mathrm{photon}\mathrm{breaks}\mathrm{one}\mathrm{covalent}\mathrm{bond}\mathrm{and}\mathrm{then}\mathrm{excites}\mathrm{two}\mathrm{H}-\mathrm{atoms}\mathrm{to}\mathrm{the}5\mathrm{th}\mathrm{orbit}. \\ \mathrm{Energy}\mathrm{of}\mathrm{photon}=\mathrm{BE}+2\times \mathrm{Excitation}\mathrm{energy} \\ =7.24\times {10}^{-19}\mathrm{J}+2\times 2.09\times {10}^{-18}\mathrm{J} \\ =4.909\times {10}^{-18}\mathrm{J} \\ \mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{4.909\times {10}^{-18}}=40.4\times {10}^{-9}\mathrm{m}=40.4\mathrm{nm} \end{gathered}$$