Problem 85 - 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 -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 320 nm light is \(9.375 \times 10^{14} \, Hz\). The energy of a mole of 320 nm photons is \(3.74 \times 10^5 \, J/mol\). Photons of UV-B radiation are more energetic than photons of UV-A radiation, making the observation that UV-B radiation is a greater cause of sunburn consistent with our findings.

Step 1: a) Frequency of 320 nm light

First, we will calculate the frequency of light having a wavelength of 320 nm. We can use the formula: \(c = \lambda v\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(v\) is the frequency. We have the value of the speed of light \(c = 3.00 \times 10^8 \, m/s\) and the wavelength, \(\lambda = 320 \, nm = 320 \times 10^{-9} \, m\). Now we can calculate the frequency (v): \(v = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \, m/s}{320 \times 10^{-9} \, m}\) \(v = 9.375 \times 10^{14} \, Hz\) The frequency of 320 nm light is \(9.375 \times 10^{14} \, Hz\).

Step 2: b) Energy of a mole of 320 nm photons

To calculate the energy of a mole of 320 nm photons, first, we need to find out the energy of a single photon using the formula: \(E = h v\), where \(E\) is the energy, \(h\) is the Planck's constant (\(6.63 \times 10^{-34} \, Js\)), and \(v\) is the frequency. From part (a), we have the frequency: \(v = 9.375 \times 10^{14} \, Hz\). Thus, the energy of a single photon can be calculated: \(E = (6.63 \times 10^{-34} \, Js)(9.375 \times 10^{14} \, Hz)\) \(E = 6.21 \times 10^{-19} \, J\) Now, we can calculate the energy of a mole of photons using the formula: \(E_{mole} = E \times N_A\), where \(N_A\) is the Avogadro's number (\(6.02 \times 10^{23} \, mol^{-1}\)). \(E_{mole} = (6.21 \times 10^{-19} \, J)(6.02 \times 10^{23} \, mol^{-1})\) \(E_{mole} = 3.74 \times 10^5 \, J/mol\) The energy of a mole of 320 nm photons is \(3.74 \times 10^5 \, J/mol\).

Step 3: c) Energy comparison between UV-A and UV-B radiation photons

We know that shorter wavelengths have higher frequencies and thus higher energy. Since UV-A radiation has wavelengths in the range of \(320-380 \, nm\), and UV-B radiation has wavelengths in the range of \(290-320 \, nm\), we can conclude that photons of UV-B radiation are more energetic than photons of UV-A radiation.

Step 4: d) Observation consistency

The observation states that UV-B radiation is considered a greater cause of sunburn in humans than UV-A radiation. In part (c), we concluded that UV-B radiation photons are more energetic than UV-A radiation photons. Thus, this observation is consistent with the answer found in part (c), as higher energy photons in UV-B radiation cause more damage to the skin, leading to sunburn.

Key Concepts

Frequency Calculation

Light spreads out in waves. A primary characteristic of these waves is their frequency. Frequency refers to the number of wave cycles passing a fixed point each second. To discover the frequency (\( v \)), when you have the wavelength (\( \lambda \)), you can use the handy formula:

• \( c = \lambda v \)

Here, \( c \) is the speed of light, a constant value (\( 3.00 \times 10^8 \, m/s \)). By rearranging the formula, you get:

• \( v = \frac{c}{\lambda} \)

For instance, if we have a wavelength of 320 nanometers (nm), it converts to 320 x 10^-9 meters when metrics standardize. Plugging this into the frequency formula, you get:

• \( v = \frac{3.00 \times 10^8 \, m/s}{320 \times 10^{-9} \, m} \)
• \( v = 9.375 \times 10^{14} \, Hz \)

This solution shows that light with a 320 nm wavelength has a high frequency of \( 9.375 \times 10^{14} \) Hertz (Hz). Understanding frequency is vital when dealing with various forms of electromagnetic radiation, including ultraviolet light.

Photon Energy

Every photon within light carries energy. This energy is a product of the frequency of the light. To determine the energy (\( E \)) of a single photon, one would use Planck's equation:

• \( E = h \cdot v \)

Planck's constant (\( h \)) is \( 6.63 \times 10^{-34} \, Js \), a fixed value essential in connecting the frequency to the energy. For light with a frequency of \( 9.375 \times 10^{14} \) Hz (computed from a 320 nm wavelength), the energy of one photon calculates to:

• \( E = (6.63 \times 10^{-34} \, Js)(9.375 \times 10^{14} \, Hz) \)
• \( E = 6.21 \times 10^{-19} \, J \)

Now, considering a mole of such photons, where Avogadro's number (\( N_A \)) is \( 6.02 \times 10^{23} \, mol^{-1} \), the total energy becomes:

• \( E_{mole} = E \times N_A \)
• \( E_{mole} = (6.21 \times 10^{-19} \, J)(6.02 \times 10^{23} \, mol^{-1}) \)
• \( E_{mole} = 3.74 \times 10^5 \, J/mol \)

This provides insight into how much energy is contained in a mole of these photons. Photon energy is central to understanding how light interacts with matter, especially regarding its potential to cause changes like sunburn.

Electromagnetic Spectrum

The electromagnetic spectrum is like a roadmap of different types of radiation organized by their wavelengths and frequencies. Each type has unique properties and effects. Ultraviolet (UV) radiation is one part of this extensive spectrum, nestled between visible light and x-rays.
The spectrum is divided into several categories:

• Radio Waves
• Microwaves
• Infrared
• Visible Light
• Ultraviolet (UV) Light
• X-rays
• Gamma Rays

UV light itself is further divided into UV-A, UV-B, and UV-C based on wavelength:

• **UV-A**: 320-380 nm - Least energetic, causes tanning.
• **UV-B**: 290-320 nm - More energetic, causes sunburn.
• **UV-C**: 100-290 nm - Most energetic, mostly stopped by Earth's atmosphere.

Understanding this spectrum helps us grasp why some kinds of radiation cause more harm than others. For example, UV radiation, with its shorter wavelengths compared to visible light, possesses higher energy levels, making it more potent in causing biological effects like sunburn.

UV-A and UV-B Comparison

While both UV-A and UV-B belong to the ultraviolet portion of the electromagnetic spectrum, they have distinct effects due to their varying wavelengths and energy levels. UV-A has longer wavelengths ranging from 320 to 380 nm, while UV-B wavelengths vary between 290 and 320 nm.
The shorter the wavelength, the higher the energy, so UV-B radiation is naturally more energetic than UV-A. This difference is significant for several reasons:

• **Energy Impact: ** UV-B photons carry more energy, making them more capable of causing skin damage such as sunburn compared to UV-A photons.
• **Health Effects: ** UV-A is primarily associated with skin aging and tanning. It penetrates more deeply into the skin layers. In contrast, UV-B is more efficient at causing direct DNA damage, hence more associated with sunburn and increased risk of skin cancer.

It's this increased energy from UV-B that elevates its risks, explaining why protective measures like sunscreen are often marketed for their UV-B blocking capabilities. The understanding of UV-A and UV-B differences also supports their target applications in medical and industrial fields.