Question

The rays of the Sun that cause tanning and burning are in the ultraviolet portion of the electromagnetic spectrum. These rays are categorized by wavelength. Socalled UV-A radiation has wavelengths in the range of $320-380 \mathrm{~nm}\(, whereas UV-B radiation has wavelengths in the range of \)290-320 \mathrm{~nm}$. (a) Calculate the frequency of light that has a wavelength of \(320 \mathrm{~nm}\). (b) Calculate the energy of a mole of 320 -nm photons. (c) Which are more energetic, photons of UV-A radiation or photons of UV-B radiation? (d) The UV-B radiation from the Sun is considered a greater cause of sunburn in humans than is UV-A radiation. Is this observation consistent with your answer to part (c)?

Short answer

The frequency of light with a wavelength of 320 nm is \(9.375 \times 10^{14} \mathrm{Hz}\). The energy of a mole of 320 nm photons is \(373.96 \mathrm{kJ/mol}\). Photons of UV-B radiation are more energetic than photons of UV-A radiation due to their shorter wavelengths. The observation that UV-B radiation causes more sunburns in humans than UV-A radiation is consistent with our findings, as the higher energy photons of UV-B radiation have a greater potential to cause harm, such as sunburns.

Step-by-step solution

(a) Calculate the frequency of 320 nm light

: Given: Wavelength, \(\lambda = 320 \mathrm{~nm} = 320 \times 10^{-9} \mathrm{~m}\) Speed of light, \(c = 3.00 \times 10^8 \mathrm{~m/s}\) To calculate the frequency, use the formula: \(v = \frac{c}{\lambda}\) \(v = \frac{3.00 \times 10^8 \mathrm{~m/s}}{320 \times 10^{-9} \mathrm{~m}}\) \(v = 9.375 \times 10^{14} \mathrm{~Hz}\) The frequency of light with a wavelength of 320 nm is \(9.375 \times 10^{14} \mathrm{~Hz}\).

(b) Calculate the energy of a mole of 320 nm photons

: Given: Frequency, \(v = 9.375 \times 10^{14} \mathrm{~Hz}\) Planck's constant, \(h = 6.63 \times 10^{-34} \mathrm{~Js}\) Avogadro's constant, \(N_A = 6.02 \times 10^{23} \mathrm{~mol^{-1}}\) First, calculate the energy of a single photon using the formula: \(E = h \times v\) \(E = 6.63 \times 10^{-34} \mathrm{~Js} \times 9.375 \times 10^{14} \mathrm{~Hz}\) \(E = 6.212 \times 10^{-19} \mathrm{~J/photon}\) Now, calculate the energy of a mole of photons by multiplying the energy of single photon by Avogadro's constant. Energy of a mole of photons = Energy of one photon × Avogadro's constant Energy of a mole of photons = \(6.212 \times 10^{-19} \mathrm{~J/photon} \times 6.02 \times 10^{23} \mathrm{~mol^{-1}}\) Energy of a mole of photons = \(373.96 \mathrm{~kJ/mol}\) The energy of a mole of 320 nm photons is \(373.96 \mathrm{~kJ/mol}\).

(c) Determine which type of radiation has more energetic photons

: Since the energy of a photon is inversely proportional to its wavelength ( \(E \propto \frac{1}{\lambda}\) ), as the wavelength decreases, the energy increases. Therefore, photons of UV-B radiation, which have shorter wavelengths (290-320 nm), are more energetic than photons of UV-A radiation (320-380 nm).

(d) Check if the observation is consistent with the answer in part (c)

: UV-B radiation causes more sunburns in humans than UV-A radiation. Our answer in part (c) shows that photons of UV-B radiation are more energetic than those of UV-A radiation. Higher energy photons have a greater potential to cause harm or damage, such as sunburns. Therefore, the observation about UV-B radiation causing more sunburn is consistent with our answer in part (c).