Octyl methoxycinnamate and oxybenzone are common ingredients in sunscreen applications. These compounds work by absorbing ultraviolet (UV) B light (wavelength 280–320 nm), the UV light most associated with sunburn symptoms. What frequency range of light do these compounds absorb?

Short Answer

Expert verified
The compounds in sunscreen, octyl methoxycinnamate and oxybenzone, absorb light with frequencies in the range of approximately \(9.375\times10^{14}\ Hz\) to \(1.071\times10^{15}\ Hz\), corresponding to the ultraviolet (UV) B light wavelength range of 280 nm to 320 nm.

Step by step solution

01

Convert the given wavelengths to meters

To make our calculations easier, we'll convert the given wavelength range of 280 nm to 320 nm into meters. Since 1 nm = 10^-9 m, we multiply each value by 10^-9 to convert them to meters. Wavelength range: 280 nm - 320 nm \(280\ nm\times10^{-9}\frac{m}{nm}=2.8\times10^{-7} m\) \(320\ nm\times10^{-9}\frac{m}{nm}=3.2\times10^{-7} m\) The converted wavelength range is now \(2.8\times10^{-7}\) m to \(3.2\times10^{-7}\) m.
02

Use the speed of light formula to find the frequency range

Now, we will use the speed of light formula to determine the frequency range related to the given wavelength range. The formula for the speed of light is: \(c = \lambda \times f\) Where c is the speed of light (3 x 10^8 m/s), λ is the wavelength, and f is the frequency. To find the frequency range, we will rearrange the formula to solve for the frequency (f): \(f = \frac{c}{\lambda}\) We will calculate the frequencies corresponding to 280 nm and 320 nm wavelengths to find the range. For the lower end of the wavelength range: \(f_1 = \frac{3\times10^8\ m/s}{2.8\times10^{-7}\ m} \approx 1.071\times10^{15}\ Hz\) For the upper end of the wavelength range: \(f_2 = \frac{3\times10^8\ m/s}{3.2\times10^{-7}\ m} \approx 9.375\times10^{14}\ Hz\)
03

Present the frequency range

Based on our calculations, the compounds in sunscreen absorb light with frequencies in the range of: Approximately \(9.375\times10^{14}\ Hz\) to \(1.071\times10^{15}\ Hz\) This is the frequency range of the ultraviolet (UV) B light that octyl methoxycinnamate and oxybenzone absorb to protect the skin from sunburn symptoms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the AgCl particles, the following reaction occurs: $$\mathrm{AgCl} \stackrel{h v}{\longrightarrow} \mathrm{Ag}+\mathrm{Cl}$$ The silver metal that is formed causes the lenses to darken. The enthalpy change for this reaction is \(3.10 \times 10^{2} \mathrm{kJ} / \mathrm{mol}\) . Assuming all this energy must be supplied by light, what is the maximum wavelength of light that can cause this reaction?

Consider the following ionization energies for aluminum: $$\begin{array}{c}{\operatorname{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} \quad I_{1}=580 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} \quad I_{2}=1815 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} \quad I_{3}=2740 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} \quad I_{4}=11,600 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between \(I_{3}\) and \(I_{4}\)

Cesium was discovered in natural mineral waters in 1860 by R. W. Bunsen and G. R. Kirchhoff using the spectroscope they invented in 1859. The name came from the Latin caesius (“sky blue”) because of the prominent blue line observed for this element at 455.5 nm. Calculate the frequency and energy of a photon of this light.

Give a possible set of values of the four quantum numbers for the 4s and 3d electrons in titanium.

Give the name and formula of each of the binary compounds formed from the following elements. a. Li and N b. Na and Br c. K and S

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free