Problem 84 - Chemistry: The Central Science

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. So-called 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 \(320-\mathrm{nm}\) 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)?

Step-by-Step Solution

Answer

The frequency of light with a wavelength of \(320 \mathrm{~nm}\) is \(9.38 \times 10^{14} \mathrm{~Hz}\). The energy of a mole of \(320-\mathrm{nm}\) photons is \(3.78 \times 10^{5}\) J/mol. Photons of UV-B radiation are more energetic than UV-A radiation, and the higher energy of UV-B radiation is consistent with its greater potential to cause sunburn in humans.

Step 1: Calculate the frequency of a \(320-\mathrm{nm}\) wavelength photon

The relationship between wavelength, frequency, and the speed of light is given by \(c = \lambda \nu\), where c is the speed of light (\(3.0 \times 10^8 \frac{\mathrm{m}}{\mathrm{s}}\)), \(\lambda\) is the wavelength, and \(\nu\) is the frequency. Given the wavelength of a 320-nm light: \( \lambda = 320 \times 10^{-9}\ \mathrm{m} \) To find the frequency, we will rearrange the formula and solve for \(\nu\): \( \nu= \frac{c}{\lambda} \)

Step 2: Calculate the energy of a single \(320-\mathrm{nm}\) photon

Now we can calculate the energy of a single photon using the relation: \( E = h \nu \), where E is the energy of the photon, h is Planck's constant (\(6.63 \times 10^{-34} \frac{\mathrm{J}\cdot\mathrm{s}}{\mathrm{photon}}\)), and \(\nu\) is the frequency. Since we have found the frequency in step one, we can calculate the energy: \( E = h \frac{c}{\lambda} \)

Step 3: Calculate the energy of a mole of \(320-\mathrm{nm}\) photons

To calculate the energy of a mole of \(320-\mathrm{nm}\) photons, we will multiply the energy of a single photon (E) with Avogadro's number (\(N_A\)): \( E_{mole} = E \cdot N_{A} \) Where \(E_{\mathrm{mole}}\) is the total energy of a mole of photons, and \(N_{A}\) is Avogadro's number (\(6.022 \times 10^{23} \mathrm{mol}^{-1}\)).

Step 4: Compare the energy of UV-A and UV-B radiation

To compare the energy of UV-A and UV-B radiation, we will observe the energy difference between their wavelengths. Since frequency and energy are directly proportional (according to the equation \(E=h \nu\)), we can compare the energies by comparing the frequencies. Therefore, photons with a shorter wavelength have higher frequencies and, in turn, higher energies.

Step 5: Discuss the relationship between the energy of UV-A, UV-B, and sunburn

The energy of UV-B radiation is higher than UV-A radiation, as shown by their shorter wavelengths. Since higher-energy photons can cause more damage to biological tissues (such as skin), we can infer that the greater energy of UV-B radiation explains their higher potential to cause sunburn in humans, which is consistent with our answer to part (c).