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Chapter 6: Problem 25

(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 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 a photon with frequency \(2.94 \times 10^{14} \mathrm{s}^{-1}\) is \(1.95 \times 10^{-19} \mathrm{J}\). (b) The energy of a photon with a wavelength of 413 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 nm.

Step by step solution

01

(a) Calculate the energy of a photon

We are given the frequency of the electromagnetic radiation: f = 2.94 × 10^14 s^-1. We will use the formula E = h * f to calculate the energy of a photon: E = (6.63 × 10^-34 Js) * (2.94 × 10^14 s^-1) = 1.95 × 10^-19 J The energy of a photon is 1.95 × 10^-19 J.
02

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

We are given the wavelength of the radiation, λ = 413 nm. In order to use the formula E = h * (c / λ), we need to convert the wavelength to meters: λ = 413 × 10^-9 m. Now, we can calculate the energy of a photon using the formula: E = (6.63 × 10^-34 Js) * (3 × 10^8 m/s) / (413 × 10^-9 m) = 4.82 × 10^-19 J The energy of a photon with a wavelength of 413 nm is 4.82 × 10^-19 J.
03

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

We are given the energy of photons: E = 6.06 × 10^-19 J. We will use the formula E = h * (c / λ) to find the wavelength of radiation: 6.06 × 10^-19 J = (6.63 × 10^-34 Js) * (3 × 10^8 m/s) / λ Now, we need to solve for λ: λ = (6.63 × 10^-34 Js) * (3 × 10^8 m/s) / (6.06 × 10^-19 J) = 3.28 × 10^-7 m We can convert the wavelength back to nanometers: λ = 328 nm. The wavelength of radiation with photons of energy 6.06 × 10^-19 J is 328 nm.

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