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Chapter 3: Problem 8

Calculate the energy of a photon whose frequency is (a) \(5 \times 10^{14} \mathrm{~Hz}\), (b) \(10 \mathrm{GHz}\), (c) \(30 \mathrm{MHz}\). Express your answers in electron volts.

Short Answer

Expert verified
(a) The energy is approximately 2.07 eV. \n(b) The energy is approximately 4.14 \times 10^{-6} eV. \n(c) The energy is approximately 1.24 \times 10^{-7} eV.

Step by step solution

01

Calculating energy in Joules

For each frequency, use the equation \( E=h \cdot f \), where \( h = 6.63 \times 10^{-34} \) Joules. seconds is the Planck constant and \( f \) is the frequency of the photon. \n(a) For \( f = 5 \times 10^{14} Hz \), \( E = h \times f = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.315 \times 10^{-19} Joules \). \n(b) For \( f = 10 GHz = 10 \times 10^9 Hz \), \( E = h \times f = 6.63 \times 10^{-34} \times 10 \times 10^9 = 6.63 \times 10^{-24} Joules \). \n(c) For \( f=30 MHz = 30 \times 10^6 Hz \), \( E = h \times f = 6.63 \times 10^{-34} \times 30 \times 10^6 = 1.99 \times 10^{-25} Joules \).
02

Converting Joules into electron volts

Since 1 eV is approximately \( 1.60 \times 10^{-19} Joules \), the conversion of Joules to eV can be done by dividing the calculated energies by \( 1.60 \times 10^{-19} \). \n(a) For \( E = 3.315 \times 10^{-19} Joules \), converting to eV gives about 2.07 eV. \n(b) For \( E = 6.63 \times 10^{-24} Joules \), converting to eV gives approximately 4.14 \times 10^{-6} eV. \n(c) For \( E = 1.99 \times 10^{-25} Joules \), converting to eV gives approximately 1.24 \times 10^{-7} eV.

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Key Concepts

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

Planck Constant
The Planck constant, represented by the symbol \( h \), is a fundamental physical constant that plays a crucial role in quantum mechanics. It connects the energy of a photon, the basic unit of light and all other forms of electromagnetic radiation, to its frequency with the simple relationship \( E = h \cdot f \). The value of the Planck constant is approximately \( 6.63 \times 10^{-34} \) Joule seconds.

This constant is not only essential for calculating photon energies but also serves as the foundation for other quantum phenomena such as the Heisenberg Uncertainty Principle. Understanding the Planck constant allows students to bridge the gap between the observed macroscopic world and the intriguing behaviors of the quantum realm.
Electron Volt Conversion
The electron volt (eV) is a unit of energy that is particularly useful in the context of atomic and subatomic processes. One electron volt is defined as the amount of energy gained (or lost) by the charge of a single electron moved across an electric potential difference of one volt. In more precise terms, 1 eV equals \(1.60 \times 10^{-19}\) Joules.

To convert the energy of a photon calculated in Joules to electron volts, you would divide the energy value by \(1.60 \times 10^{-19}\). For example, a photon energy of \(3.315 \times 10^{-19}\) Joules is roughly equivalent to 2.07 eV. The usefulness of this conversion lies in its ability to provide energy values in a scale that is more relatable to the energies typically associated with electrons in atoms, making it easier to understand and compare with other quantum phenomena.
Photon Frequency
Photon frequency, denoted by \( f \), refers to the number of wave cycles of a photon that pass a fixed point per second. It is typically measured in Hertz (Hz), with one Hertz equating to one cycle per second. Photon frequency directly dictates the energy of a photon through the relationship \( E = h \cdot f \), where \( E \) represents the photon's energy and \( h \) the Planck constant.

Frequencies can vary across a wide spectrum of values, with visible light falling in the range of about \( 4 \times 10^{14} \) to \( 7 \times 10^{14} \) Hz. In the exercises provided, we see examples like \( 5 \times 10^{14} \) Hz representing a high energy visible light photon, while lower frequencies like \( 10 \) GHz and \( 30 \) MHz correspond to radiofrequencies, which carry much less energy per photon. Understanding how frequency affects energy helps to explain why different types of electromagnetic radiation interact with matter in diverse ways.

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Most popular questions from this chapter

X-rays with an energy of \(300 \mathrm{keV}\) undergo Compton scattering from a target. If the scattered rays are detected at \(30^{\circ}\) relative to the incident rays, find (a) the Compton shift at this angle, (b) the energy of the scattered x-ray, and (c) the energy of the recoiling electron.

X-radiation from a molybdenum target \((0.626 \AA)\) is incident on a crystal with adjacent atomic planes spaced \(4.00 \times 10^{-10} \mathrm{~m}\) apart. Find the three smallest angles at which intensity maxima occur in the diffracted beam.

Planck's fundamental constant, \(h\). Planck ultimately realized the great and fundamental importance of \(h\), which, much more than a curve-fitting parameter, is actually the measure of all quantum phenomena. In fact, Planck suggested using the universal constants \(h, c\) (the velocity of light), and \(G\) (Newton's gravitational constant) to construct "natural" or universal units of length, time, and mass. He reasoned that the current units of length, time, and mass were based on the accidental size, motion, and mass of our particular planet, but that truly universal units should be based on the quantum theory, the speed of light in a vacuum, and the law of gravitation - all of which hold anywhere in the universe and at all times. Show that the expressions \(\left(\frac{h G}{c^{3}}\right)^{1 / 2},\left(\frac{h G}{c^{5}}\right)^{1 / 2}\), and \(\left(\frac{h c}{G}\right)^{1 / 2}\) have dimensions of length, time, and mass and find their numerical values. These quantities are called, respectively, the Planck length, the Planck time, and the Planck mass. Would you care to speculate on the physical meaning of these quantities?

The average power generated by the Sun has the value \(3.74 \times 10^{26} \mathrm{~W}\). Assuming the average wavelength of the Sun's radiation to be \(500 \mathrm{~nm}\), find the number of photons emitted by the Sun in \(1 \mathrm{~s}\).

Consider the metals lithium, beryllium, and mercury, which have work functions of \(2.3 \mathrm{eV}, 3.9 \mathrm{eV}\), and \(4.5 \mathrm{eV}\), respectively. If light of wavelength \(300 \mathrm{~nm}\) is incident on each of these metals, determine (a) which metals exhibit the photoelectric effect and (b) the maximum kinetic energy for the photoelectron in each case.
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