What will be the energy in kJ/mol associated with a radio wave of wavelength equal to \(3 \text{ × }10^{13}\overset{o}{A}\)?

\(3.990\text{ × }10^{-8}kJ/mol\)

The correct answer is option 4.\(3.990\text{ × }10^{-8}kJ/mol\).

\(\text{We know that, }\)

\(\text{E = h}\nu \text{ = h}\frac{c}{\lambda}\)

\(\text{Now, since h is in Js, c is in m/s and }\lambda \text{is in }\overset{o}{A}\text{, E cannot be obtained in joules. Thus in order to get E in joules, }\lambda \text{ should be converted to m. Thus:}\)

\(\lambda \text{ = 3 × }10^{13}\overset{o}{A}\text{ = 3 × }10^{13}\text{ × }10^{-10}m\text{ = 3 × }10^3m\)

\(\text{c = 3 × }10^8ms^{-1}\)

\(\text{h = 6.626 × }10^{-34}Js\)

∴ \(\text{E = }\frac{(6.626\text{ × }10^{-34}Js)\text{ × }(3\text{ × }10^8ms^{-1})}{3\text{ × }10^3}m\)

\(\text{  E = 6.626 × }10^{-29}J/atom\)

\(\text{Now since Avogadro's number = 6.023 × }10^{23}\)

∴ \(\text{E = 6.626 × }10^{-29}\text{ × 6.023 × }10^{23}\text{ J}/mole\)

\(\text{        = }\frac{\text{6.626 × }10^{-29}\text{ × 6.023 × }10^{23}}{10^3}\text{ kJ}/mol \)

\(\text{        =  6.626 × 6.023 × }10^{-9}\text{ kJ}/mol\)

\(\text{          =  3.990 × }10^{-8}\text{ kJ}/mol\)