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Given that the wavelength of light, $\\lambda$, is equal to {wavelength_mantissa} $\\times$ 10{wavelength_log} m;

", "advice": "

\n

a) If you are given a quantity in wavenumbers, convert this into a wavelength;

\n

\\[[\\frac{1}{\\lambda}={\\rm wavenumber}]~{\\rm and~therefore}~[\\frac{1}{\\rm wavenumber}={\\lambda}]\\]

\n

If you specify wavelength in cm, you will obtain the wavenumber in cm-1

\n

\\[\\frac{1}{{\\var{wavelength_mantissa}~\\times~10^{\\var{wavelength_log_cm}}~\\rm cm}}=\\var{wavenumber}~\\rm cm^{-1}\\]

\n

\n

\n

\n

b) Calculate the energy of a single photon using; 

\n

\\[E=h\\nu=6.626~\\times~10^{-34}~{\\rm~J~s}~\\times~\\var{Frequency_mantissa}~\\times~10^{\\var{Frequency_log}}~{\\rm~s^{-1}}=\\var{Energy_mantissa}\\times~10^\\var{Energy_log}~{\\rm~J}\\]

\n

\n

c) Then, multiply by the Avogadro number, and divide by 1000 J kJ-1 to convert this into an energy in kJ mol-1

\n

\\[\\frac{\\var{Energy_mantissa}\\times~10^\\var{Energy_log}~{\\rm~J}~\\times~6.022~\\times~10^{23}~{\\rm~mol^{-1}}}{1000~\\rm J~kJ^{-1}}=\\var{Energy_molar}~{\\rm kJ~mol^{-1}}\\]

\n

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randomi

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what is the wavenumber in units of cm-1

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The energy, E, of a single photon is; 

\n

[[0]] $\\times$ 10[[1]] J

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The energy, E, of one mole of photons is [[0]] kJ mol-1.

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