Question

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{~nm}\). (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? (d) These UV photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the 325-nm radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

Short answer

(a) The energy of a photon of 325 nm wavelength is approximately \(6.11 \times 10^{-19} \mathrm{J}\). (b) The energy of one mole of these photons is approximately \(367.6 \mathrm{kJ/mol}\). (c) There are approximately \(1.64 \times 10^{15}\) photons in a \(1.00 \mathrm{mJ}\) burst of this radiation. (d) The average energy of chemical bonds in skin is estimated at \(367.6 \mathrm{kJ/mol}\).

Step-by-step solution

Calculate the frequency of the photon

Use the speed of light equation to find the frequency, remembering to convert the wavelength from nm to meters: \(c = \lambda \times f\) \(f = \frac{c}{\lambda}\) \(f = \frac{3.00 \times 10^8 \mathrm{m/s}}{325 \times 10^{-9} \mathrm{m}}\) \(f = 9.23 \times 10^{14} \mathrm{Hz}\)

Determine the energy of a single photon

Use Planck's equation and the frequency obtained in Step 1: \(E = h \times f\) \(E = (6.63 \times 10^{-34} \mathrm{Js}) \times (9.23 \times 10^{14} \mathrm{Hz})\) \(E = 6.11 \times 10^{-19} \mathrm{J}\)

Calculate the energy of one mole of photons

Multiply the energy of a single photon by Avogadro's number: \(E_{mole} = E \times N_A\) \(E_{mole} = (6.11 \times 10^{-19} \mathrm{J}) \times (6.022 \times 10^{23} \mathrm{mol^{-1}})\) \(E_{mole} = 367.6 \mathrm{kJ/mol}\)

Calculate the number of photons in a 1.00 mJ energy burst

Divide the energy burst by the energy of a single photon, remembering to convert mJ to J: \(n = \frac{1.00 \times 10^{-3} \mathrm{J}}{6.11 \times 10^{-19} \mathrm{J}}\) \(n \approx 1.64 \times 10^{15} \mathrm{photons}\)

Estimate the average energy of chemical bonds in skin

From the information given in the exercise, the energy of a single photon is enough to break one chemical bond in the skin. Using the energy per mole of photons from Step 3, we can estimate the average energy of these chemical bonds: \(E_{bond} \approx 367.6 \mathrm{kJ/mol}\) In summary: (a) The energy of a photon of 325 nm wavelength is approximately \(6.11 \times 10^{-19} \mathrm{J}\). (b) The energy of one mole of these photons is approximately \(367.6 \mathrm{kJ/mol}\). (c) There are approximately \(1.64 \times 10^{15}\) photons in a \(1.00 \mathrm{mJ}\) burst of this radiation. (d) The average energy of chemical bonds in skin is estimated at \(367.6 \mathrm{kJ/mol}\).