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(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{~s}^{-1}\). (b) Calculate the energy of a photon of radiation whose wavelength is \(413 \mathrm{~nm}\). (c) What wavelength of radiation has photons of energy \(6.06 \times 10^{-19} \mathrm{~J}\) ?

Short Answer

Expert verified
(a) The energy of the photon with a frequency of \(2.94 \times 10^{14} \mathrm{~s}^{-1}\) is \(1.95 \times 10^{-19} \mathrm{~J}\). (b) The energy of the photon with a wavelength of \(413 \mathrm{~nm}\) is \(4.82 \times 10^{-19} \mathrm{~J}\). (c) The wavelength of radiation with photons of energy \(6.06 \times 10^{-19} \mathrm{~J}\) is \(328 \mathrm{~nm}\).

Step by step solution

01

(a) Calculate the energy of a photon with given frequency

To calculate the energy of the photon, use the Planck's formula: E = hν. Given frequency, ν = 2.94 x 10^14 s⁻¹ Planck's constant, h = 6.626 x 10⁻³⁴ Js So, E = (6.626 x 10^-34 Js) * (2.94 x 10^14 s⁻¹) E ≈ 1.95 x 10^-19 J Thus, the energy of the photon is 1.95 x 10^-19 J.
02

(b) Calculate the energy of a photon with given wavelength

To calculate the energy of the photon, we first need to find its frequency (ν) using the speed of light formula: c = λν. Given wavelength, λ = 413 nm = 413 x 10⁻⁹ m Speed of light, c = 3.0 x 10^8 m/s Now, rearrange the formula for frequency: ν = c / λ ν = (3.0 x 10^8 m/s) / (413 x 10⁻⁹ m) ν ≈ 7.27 x 10^14 s⁻¹ Now we have the frequency, we can find the energy using the Planck's formula: E = hν. E = (6.626 x 10^-34 Js) * (7.27 x 10^14 s⁻¹) E ≈ 4.82 x 10^-19 J Thus, the energy of the photon is 4.82 x 10^-19 J.
03

(c) Find the wavelength of radiation with given photon energy

To find the wavelength of radiation, we first need to find the frequency (ν) of the photons using the Planck's formula: E = hν. Given energy, E = 6.06 x 10^-19 J Planck's constant, h = 6.626 x 10⁻³⁴ Js Now, rearrange the formula for frequency: ν = E / h ν = (6.06 x 10^-19 J) / (6.626 x 10^-34 Js) ν ≈ 9.15 x 10^14 s⁻¹ Now we have the frequency, we can find the wavelength using the speed of light formula: c = λν. Speed of light, c = 3.0 x 10^8 m/s Now, rearrange the formula for wavelength: λ = c / ν λ = (3.0 x 10^8 m/s) / (9.15 x 10^14 s⁻¹) λ ≈ 3.28 x 10^-7 m Convert the wavelength to nm: λ ≈ 328 nm Thus, the wavelength of radiation with photons of energy 6.06 x 10⁻¹⁹ J is 328 nm.

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Key Concepts

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

Planck's formula
Understanding the relationship between energy and frequency in electromagnetic radiation begins with Planck's formula, which is the cornerstone of quantum mechanics. This fundamental equation is expressed as E = h\(u\), where E represents the energy of a photon, h is Planck’s constant (approximately \(6.626 \times 10^{-34} \) Joule-seconds), and \(u\) is the frequency of the radiation.

When observing the behavior of photons, tiny packets of energy, Planck's formula demonstrates that the more frequent the oscillation of electromagnetic waves, the higher the energy carried by the photons. This concept is further highlighted through exercises that prompt students to calculate photon energy, providing a practical understanding of this quantum relationship. Planck's constant provides a conversion factor between the wavelength and the energy, bridging the gap between the macroscopic world we observe and the microscopic realm of quantum particles.
Frequency of electromagnetic radiation
Electromagnetic radiation covers a broad range of frequencies, from radio waves to gamma rays, and it is essentially light, including wavelengths not visible to the human eye. The frequency of this radiation, denoted by \(u\), is the number of times the wave oscillates or cycles per second and is measured in Hertz (Hz), signifying cycles per second or s−1.

Higher frequencies correlate with higher energy photons, as per Planck's formula. For example, gamma rays have frequencies at the high end of the spectrum, and consequently, they possess high photon energies, which is why they can be so penetrative and potentially harmful. When estimating the energy of a photon for a given frequency, as demonstrated in exercise solutions, Planck's formula allows us to directly calculate this energy without needing to consider the photon's wavelength.
Wavelength and frequency relationship
The wavelength (\(\lambda\)) and frequency (\(u\)) of electromagnetic radiation are interconnected through the speed of light with the equation \(c = \lambdau\), where \(c\) stands for the speed of light in a vacuum, which is about \(3.0 \times 10^8 \) meters per second. This inverse relationship means that as the wavelength increases (representing the physical distance between successive crests of the wave), the frequency decreases, and vice versa.

Speed of Light as a Constant

In lessons on electromagnetic radiation, we stress that the speed of light is a constant in a vacuum, making it a pivotal factor in calculations involving wavelengths and frequencies. When the wavelength of a photon is given, like in the provided exercise, this unwavering speed permits the calculation of the unknown frequency or energy of the photon, emphasizing the interconnected nature of these fundamental properties of light.
Speed of light
The speed of light (\(c\)) is a fundamental constant of nature, approximately \(299,792,458 \) meters per second in a vacuum. Its constancy enables precise calculations in various astrophysical and physics problems. In optics, as well as in the provided textbook exercises, the speed of light is instrumental for deriving other quantities such as frequency and wavelength of electromagnetic waves.

One significant implication of the speed of light is that it establishes a speed limit for how fast information, or anything else, can travel in the universe. This speed limit, along with Planck's constant, forms the basis of our ability to understand the scale and energy of various elements in the quantum world, including the photons whose energies students calculate in exercises.

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Most popular questions from this chapter

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitted because sodium ions in the pickle become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(589 \mathrm{~nm}\). Calculate its frequency. (b) What is the energy of \(0.10 \mathrm{~mol}\) of these photons? (c) Calculate the energy gap between the excited and ground states for the sodium ion. (d) If you soaked the pickle for a long time in a different salt solution, such as strontium chloride, would you still observe \(589-\mathrm{nm}\) light emission?

The discovery of hafnium, element number 72 , provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements 58-71) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation-reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2\). Write a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

The Lyman series of emission lines of the hydrogen atom are those for which \(n_{\mathrm{f}}=1\). (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series-those for which \(n_{\mathrm{i}}=2,3\), and 4 .

The Chemistry and Life box in Section 6.7 described the techniques called NMR and MRI. (a) Instruments for obtaining MRI data are typically labeled with a frequency, such as 600 MHz. In what region of the electromagnetic spectrum does a photon with this frequency belong? (b) What is the value of \(\Delta E\) in Figure 6.27 that would correspond to the absorption of a photon of radiation with frequency \(450 \mathrm{MHz} ?(\mathbf{c})\) When the 450 -MHz photon is absorbed, does it change the spin of the electron or the proton on a hydrogen atom?

Calculate the uncertainty in the position of (a) an electron moving at a speed of \((3.00 \pm 0.01) \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathbf{b})\) a neutron moving at this same speed. (The masses of an electron and a neutron are given in the table of fundamental constants in the inside cover of the text.) (c) Based on your answers to parts (a) and (b), which can we know with greater precision, the position of the electron or of the neutron?

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