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Chapter 42: Problem 2

During each of these processes, a photon of light is given up. In each process, what wavelength of light is given up, and in what part of the electromagnetic spectrum is that wavelength? (a) A molecule decreases its vibrational energy by \(0.198 \mathrm{eV} ;\) (b) an atom decreases its energy by \(7.80 \mathrm{eV} ;\) (c) a molecule decreases its rotational energy by \(4.80 \times 10^{-3} \mathrm{eV}\)

Short Answer

Expert verified
The wavelengths of light emitted due to the energy decreases are calculated using the Planck-Einstein relation. The resulting wavelengths are then used to identify the corresponding parts of the electromagnetic spectrum.

Step by step solution

01

Convert energy from eV to Joules

We first have to convert the given energy values from electron-volts (eV) to Joules (J) to be consistent with SI units used in the Planck-Einstein relation. We use the conversion factor : \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\) to carry out the conversion.
02

Identify constants

Identify and record the values of the two constants in the Planck-Einstein equation: Planck's constant, \(h = 6.63 \times 10^{-34}\) J鈰卻, and the speed of light, \(c = 3 \times 10^8\) m/s.
03

Calculate the wavelength

Use the rearranged Planck-Einstein equation \(\lambda = h \cdot c / E \) to calculate the wavelength of the light emitted due to the energy decrease in each process.
04

Identify the part of the electromagnetic spectrum

Identify the part of the electromagnetic spectrum that the calculated wavelengths correspond to. Here's a general guide: Radio > 1m, Microwave 1m to 1mm, Infrared 1mm to 700nm, Visible 700nm to 400nm, Ultraviolet 400nm to 10nm, X-rays 10nm to 0.01nm and Gamma rays < 0.01nm

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Key Concepts

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

Photon Energy
Every time a photon is involved, energy becomes an exciting topic! In our case, a photon is emitted during a process where energy is given up, such as a molecule decreasing its energy. The energy of a photon can be thought of as a small package of energy. This energy is measured in units called electron-volts (eV).
- In this exercise, photon energy changes due to processes like vibrational, atomic, or rotational transitions - When a molecule or atom gives up energy, it releases this energy as a photon - The measure of energy in eV shows how much energy is contained in that photon
Whenever you are working with energy involving photons, whether calculating or converting, equip yourself with the proper units to ensure accuracy.
Planck-Einstein Relation
The Planck-Einstein relation is a fundamental tool that connects the energy of a photon to its wavelength. This relationship is expressed in the equation:- \[ E = h \cdot f \] where:
  • \( E \) is the energy of the photon in Joules
  • \( h \) is Planck's constant (6.63 \times 10^{-34} J鈰卻)
  • \( f \) is the frequency of the light in Hertz
This equation can be rearranged to express wavelength (\( \lambda \) ) based on the energy, as used in our exercise:
  • \( \lambda = \frac{h \cdot c}{E} \)
  • Where \( c \) is the constant speed of light (3 \times 10^8 m/s)
Understanding this relation helps in calculating how energy levels result in specific wavelengths of emitted light.
Wavelength Calculation
Wavelength calculation is essential to determine which part of the electromagnetic spectrum is involved. For different energy releases, the wavelength is found using the equation:
\( \lambda = \frac{h \cdot c}{E} \)- Start by converting the given energy from electron-volts (eV) to Joules (J)using the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19}\text{ J}\)- Then, insert these energy values, along with Planck's constant (h) and the speed of light (c), into the rearranged equation- Solve for the wavelength \( \lambda \), which will be in meters
Each calculated wavelength lets us pinpoint exactly where that light falls on the electromagnetic spectrum! This method is reliable and leads to a precise understanding of energy behavior.
Electromagnetic Spectrum Regions
The electromagnetic spectrum is a vast range of wavelengths that light can have, extending from long, low-energy waves to short, high-energy waves.
- Radio waves: > 1m - Microwaves: 1m to 1mm - Infrared: 1mm to 700nm - Visible light: 700nm to 400nm - Ultraviolet: 400nm to 10nm - X-rays: 10nm to 0.01nm - Gamma rays: < 0.01nm
By measuring the wavelength of light given up by molecules or atoms, you can determine its position on the electromagnetic spectrum. Different parts of the spectrum have various applications and characteristics, such as visible light enabling us to see and microwaves used in cooking. Identifying where light falls within this spectrum reveals a lot about its energy and potential uses.

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Most popular questions from this chapter

Consider the \(\mathrm{CO}_{2}\) molecule shown in Fig. \(42.10 \mathrm{c} .\) The oxygen molecules have mass \(M_{\mathrm{O}}\) and the carbon atom has mass \(M_{\mathrm{C}}\). Parameterize the positions of the left oxygen atom, the carbon atom, and the right oxygen atom using \(x_{1}, x_{2},\) and \(x_{3}\) as the respective rightward deviations from equilibrium. Treat the bonds as Hooke's-law springs with common spring constant \(k^{\prime}\). (a) Use Newton's second law to obtain expressions for \(M_{i} \ddot{x}_{i}=M_{i} d^{2} x / d t^{2}\) in each case \(i=1,2,3,\) where \(M_{1,2,3}=\left(M_{\mathrm{O}}, M_{\mathrm{C}}, M_{\mathrm{O}}\right) .\) (Note: We represent time derivatives using dots.) Assume \(X_{\mathrm{C}}=0\) at \(t=0 .\) (b) To ascertain the motion of the asymmetric stretching mode, set \(x_{1}=x_{3} \equiv X_{0}\) and set \(x_{2} \equiv X_{\mathrm{C}}\). Write the two independent equations that remain from your previous result. (c) Eliminate the sum \(X_{\mathrm{O}}+X_{\mathrm{C}}\) from your equations. Use what remains to ascertain \(X_{\mathrm{C}}\) in terms of \(X_{\mathrm{O}}\). (d) Substitute your expression for \(X_{\mathrm{C}}\) into your equation for \(X_{\mathrm{O}}\) to derive a harmonic oscillator equation \(M_{\mathrm{eff}} \ddot{X}_{\mathrm{O}}=-k \dot{X}_{\mathrm{O}} .\) What is \(M_{\mathrm{eff}} ?\) (e) This equation has the solution \(X_{\mathrm{O}}(t)=A \cos (\omega t) .\) What is the angular frequency \(\omega ?\) (f) Using the experimentally determined spring constant \(k^{\prime}=1860 \mathrm{~N} / \mathrm{m}\) and the atomic masses \(M_{\mathrm{C}}=12 \mathrm{u}\) and \(M_{\mathrm{O}}=16 \mathrm{u},\) where \(\mathrm{u}=1.6605 \times 10^{-27} \mathrm{~kg},\) to determine the oscillation frequency \(f=\omega / 2 \pi\)

Consider a system of \(N\) free electrons within a volume \(V\). Even at absolute zero, such a system exerts a pressure \(p\) on its surroundings due to the motion of the electrons. To calculate this pressure, imagine that the volume increases by a small amount \(d V\). The electrons will do an amount of work \(p d V\) on their surroundings, which means that the total energy \(E_{\text {tot }}\) of the electrons will change by an amount \(d E_{\mathrm{tot}}=-p d V .\) Hence \(p=-d E_{\mathrm{tot}} / d V\) (a) Show that the pressure of the electrons at absolute zero is $$ p=\frac{3^{2 / 3} \pi^{4 / 3} \hbar^{2}}{5 m}\left(\frac{N}{V}\right)^{5 / 3} $$ (b) Evaluate this pressure for copper, which has a free-electron concentration of \(8.45 \times 10^{28} \mathrm{~m}^{-3} .\) Express your result in pascals and in atmospheres. (c) The pressure you found in part (b) is extremely high. Why, then, don't the electrons in a piece of copper simply explode out of the metal?

If a sodium chloride \((\mathrm{NaCl})\) molecule could undergo an \(n \rightarrow n-1\) vibrational transition with no change in rotational quantum number, a photon with wavelength \(20.0 \mu \mathrm{m}\) would be emitted. The mass of a sodium atom is \(3.82 \times 10^{-26} \mathrm{~kg},\) and the mass of a chlorine atom is \(5.81 \times 10^{-26} \mathrm{~kg}\). Calculate the force constant \(k^{\prime}\) for the interatomic force in \(\mathrm{NaCl}\).

Two atoms of cesium (Cs) can form a Cs \(_{2}\) molecule. The equilibrium distance between the nuclei in a \(\mathrm{Cs}_{2}\) molecule is \(0.447 \mathrm{nm}\). Calculate the moment of inertia about an axis through the center of mass of the two nuclei and perpendicular to the line joining them. The mass of a cesium atom is \(2.21 \times 10^{-25} \mathrm{~kg}\).

The table gives the occupation probabilities \(f(E)\) as a function of the energy \(E\) for a solid conductor at a fixed temperature \(T\). $$ \begin{array}{l|cccccc} f(\boldsymbol{E}) & 0.064 & 0.173 & 0.390 & 0.661 & 0.856 & 0.950 \\ \hline \boldsymbol{E}(\mathrm{eV}) & 3.0 & 2.5 & 2.0 & 1.5 & 1.0 & 0.5 \end{array} $$ To determine the Fermi energy of the solid material, you are asked to analyze this information in terms of the Fermi-Dirac distribution. (a) Graph the values in the table as \(E\) versus \(\ln \\{[1 / f(E)]-1\\} .\) Find the slope and \(y\) -intercept of the best-fit straight line for the data points when they are plotted this way. (b) Use your results of part (a) to calculate the temperature \(T\) and the Fermi energy of the material.
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