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Chapter 42: Problem 3

To break a chemical bond in the molecules of human skin and thus cause sunburn, a photon energy of about \(3.50 \mathrm{eV}\) is required. To what wavelength does this correspond? $$ \lambda=\frac{h \mathrm{c}}{\mathrm{E}}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{(3.50 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}=354 \mathrm{~nm} $$ Ultraviolet radiation causes sunburn.

Short Answer

Expert verified
The corresponding wavelength is 354 nm, which is in the ultraviolet range.

Step by step solution

01

Understand the Energy-Wavelength Relationship

The wavelength \( \lambda \) of a photon is related to its energy \( E \) by the equation: \( \lambda = \frac{hc}{E} \), where \( h \) is Planck's constant \( 6.63 \times 10^{-34} \ \mathrm{J \cdot s} \) and \( c \) is the speed of light \( 2.998 \times 10^8 \ \mathrm{m/s} \).
02

Convert Photon Energy to Joules

The photon's energy is given as \( 3.50 \ \mathrm{eV} \). To use it in our wavelength formula, convert this energy to Joules using the conversion factor \( 1 \ \mathrm{eV} = 1.602 \times 10^{-19} \ \mathrm{J} \). Thus, \( 3.50 \ \mathrm{eV} \) is equal to \( 3.50 \times 1.602 \times 10^{-19} \ \mathrm{J} \).
03

Calculate Wavelength \( \lambda \)

Substitute the values into the wavelength formula: \( \lambda = \frac{(6.63 \times 10^{-34} \ \mathrm{J \cdot s})(2.998 \times 10^8 \ \mathrm{m/s})}{3.50 \times 1.602 \times 10^{-19} \ \mathrm{J}} \). Simplify the fraction to find \( \lambda \).
04

Simplifying to Find the Wavelength

Calculate the value of \( \lambda \): \( \lambda = \frac{6.63 \times 10^{-34} \times 2.998 \times 10^8}{3.50 \times 1.602 \times 10^{-19}} = 354 \ \mathrm{nm} \).
05

Interpretation

The wavelength \( 354 \ \mathrm{nm} \) indicates that this photon is in the ultraviolet region of the electromagnetic spectrum, which is known to cause sunburn.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Energy
Photon energy refers to the amount of energy carried by a single photon, the smallest unit of light. Understanding photon energy involves a simple equation: \( E = h u \), where \( E \) is the energy, \( h \) is Planck's constant \( (6.63 \times 10^{-34} \, \mathrm{J \cdot s}) \), and \( u \) is the frequency of the photon.
This equation reveals that energy is directly proportional to frequency.
  • High-frequency photons (such as gamma rays) have high energy.
  • Low-frequency photons (like radio waves) have low energy.
To calculate the wavelength corresponding to a specific photon energy in electron volts (eV), you need to convert eV into Joules using the conversion \( 1 \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J} \).
This helps in utilizing the wavelength formula \( \lambda = \frac{hc}{E} \), allowing us to find which part of the electromagnetic spectrum the photon belongs to.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation. Radiation is the emission of energy waves, which can travel through space.
This spectrum is divided into sections based on wavelength and frequency.
  • Radio waves have the longest wavelengths and the lowest frequencies.
  • Gamma rays have the shortest wavelengths and the highest frequencies.
Understanding that frequency is inversely proportional to wavelength is critical.
As the equation \( c = \lambda u \) shows, where \( c \) is the speed of light (\( 2.998 \times 10^8 \, \mathrm{m/s} \)), an increase in frequency means a decrease in wavelength.
This relationship explains why high energy photons, like ultraviolet (UV) rays, have shorter wavelengths.
Each part of the spectrum corresponds to various energy levels that affect matter differently, such as visible light enabling us to see and ultraviolet light causing sunburn.
Ultraviolet Radiation
Ultraviolet (UV) radiation is a type of electromagnetic radiation with wavelengths between approximately 10 nm and 400 nm.
Placed right next to visible light in the spectrum, UV rays hold more energy than visible light and less than X-rays.
  • UV is further divided into UVA, UVB, and UVC, with UVC being the most energetic.
  • Only UVA and UVB reach the Earth's surface because the atmosphere absorbs UVC.
UV radiation is notorious for having high energy that can cause damage to the skin, leading to sunburns and increasing the risk of skin cancer.
In our wavelength calculation, a photon energy of \( 3.50 \, \mathrm{eV} \) corresponds to a wavelength of 354 nm, placing it in the UV range.
Protecting from excessive UV exposure is essential to limit harmful effects, using sunscreen or protective clothing.

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Most popular questions from this chapter

What is the de Broglie wavelength for a particle moving with speed \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) if the particle is \((a)\) an electron, \((b)\) a proton, and \((c)\) a \(0.20\) -kg ball? We make use of the definition of the de Broglie wavelength: $$ \lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{m\left(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}=\frac{3.31 \times 10^{-40} \mathrm{~m} \cdot \mathrm{kg}}{m} $$ Substituting the required values for \(m\), one finds that the wavelength is \(3.6 \times 10^{-10} \mathrm{~m}\) for the electron, \(2.0 \times 10^{-13} \mathrm{~m}\) for the proton, and \(1.7 \times 10^{-39} \mathrm{~m}\) for the \(0.20\) -kg ball.

II] A particle of mass \(m\) is confined to a circular orbit with radius \(R\). For resonance of its de Broglie wave on this orbit, what energies can the particle have? Determine the KE for an electron with \(R=\) \(0.50 \mathrm{~nm}\). To resonate on a circular orbit, a wave must circle back on itself in such a way that crest falls upon crest and trough falls upon trough. One resonance possibility (for an orbit circumference that is four wavelengths long) is shown in \(\underline{\text { Fig. } 42-2 . \text { In general, resonance }}\) occurs when the circumference is \(n\) wavelengths long, where \(n=\) \(1,2,3, \ldots .\) For such a de Broglie wave $$ n \lambda_{n}=2 \pi R \quad \text { and } \quad p_{n}=\frac{h}{\lambda_{n}}=\frac{n h}{2 \pi R} $$ Fig. \(42-2\) As in Problem \(42.17\), $$ (\mathrm{KE})_{n}=\frac{p_{n}^{2}}{2 m}=\frac{n^{2} h^{2}}{8 \pi^{2} R^{2} m} $$ The energies are obviously quantized. Placing in the values requested leads to $$ (\mathrm{KE})_{n}=2.4 \times 10^{-20} n^{2} \mathrm{~J}=0.15 n^{2} \mathrm{eV} $$

The work function of sodium metal is \(2.3 \mathrm{eV}\). What is the longestwavelength light that can cause photoelectron emission from sodium? At threshold, the photon energy just equals the energy required to tear the electron loose from the metal. In other words, the electron's \(\mathrm{KE}\) is zero and so \(h f=\varphi\). Since \(f=c / \lambda\) $$ \begin{array}{c} \phi=\frac{h \mathrm{c}}{\lambda} \\ (2.3 \mathrm{eV})\left(\frac{\left(1.602 \times 10^{-19} \mathrm{~J}\right.}{1.00 \mathrm{eV}}\right)=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{\lambda} \\ \lambda=5.4 \times 10^{-7} \mathrm{~m} \end{array} $$

What are the speed and momentum of a 500 -nm photon?

An X-ray beam with a wavelength of exactly \(5.00 \times 10^{-14} \mathrm{~m}\) strikes a proton that is at rest \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\). If the X-rays are scattered through an angle of \(110^{\circ}\), what is the wavelength of the scattered X-rays?
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