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Chapter 27: Problem 57

A muon and an antimuon, each with a mass that is 207 times greater than an electron, were at rest when they annihilated and produced two photons of equal energy. What is the wavelength of each of the photons?

Short Answer

Expert verified
Answer: To find the wavelength of the photons produced when a muon and an antimuon annihilate, we need to follow these steps: 1. Calculate the mass of the muon and antimuon (207 times the mass of an electron). 2. Calculate their energy using E=mc^2. 3. Use energy conservation to find the energy of the photons. 4. Find the energy of a single photon (half the total energy). 5. Use the energy-wavelength relationship, E=hc/λ, to find their wavelength. 6. Plug in the numerical values to calculate the wavelength of the photons. The final equation for the wavelength will be λ = (6.63 × 10^-34 J*s)(3 × 10^8 m/s) / E_photon.

Step by step solution

01

Find the mass of the muon and antimuon

First, let's find the mass of the muon and antimuon using the given information. Since the mass of each is 207 times greater than the mass of an electron, their mass can be calculated as follows: Mass of muon (m) = Mass of antimuon (m') = 207 * electron_mass where electron_mass = 9.11 × 10^-31 kg
02

Calculate the energy of muon and antimuon using E=mc^2

Now we can calculate the energy of the muon and the antimuon using the energy-mass equivalence formula: E = mc^2 E' = m'c^2 where E and E' are the energies of the muon and antimuon, respectively, and c is the speed of light (3 × 10^8 m/s).
03

Conserve the energy when the particles annihilate

When the muon and antimuon annihilate, their energy is conserved and transformed into the energy of the two photons. Since the photons have equal energy, we can express the conservation of energy as: E_total = E + E' = 2 * E_photon where E_total is the total energy of muon and antimuon, and E_photon is the energy of one photon.
04

Find energy of a single photon

From the previous equation, we can determine the energy of a single photon: E_photon = (E + E') / 2 E_photon = E_total / 2
05

Find the wavelength of the photons

Now that we have the energy of a single photon, we can use the energy-wavelength relationship to find the wavelength of the photons: E_photon = hc / λ where h is Planck's constant (6.63 × 10^-34 J*s), and λ is the wavelength of one photon. Solving for the wavelength, we get: λ = hc / E_photon
06

Calculate the wavelength of the photons

Finally, plug in the values for h, c, and E_photon into the equation for λ to find the wavelength of the photons: λ = (6.63 × 10^-34 J*s)(3 × 10^8 m/s) / E_photon Calculate the result, and you will have the wavelength of each of the photons produced when a muon and an antimuon annihilate.

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Most popular questions from this chapter

An owl has good night vision because its eyes can detect a light intensity as faint as \(5.0 \times 10^{-13} \mathrm{W} / \mathrm{m}^{2} .\) What is the minimum number of photons per second that an owl eye can detect if its pupil has a diameter of \(8.5 \mathrm{mm}\) and the light has a wavelength of $510 \mathrm{nm} ?$
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In gamma-ray astronomy, the existence of positrons \(\left(\mathrm{e}^{+}\right)\) can be inferred by a characteristic gamma ray that is emitted when a positron and an electron (e \(^{-}\) ) annihilate. For simplicity, assume that the electron and positron are at rest with respect to an Earth observer when they annihilate and that nothing else is in the vicinity. (a) Consider the reactions $\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow \gamma$, where the annihilation of the two particles at rest produces one photon, and \(\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow 2 \gamma,\) where the annihilation produces two photons. Explain why the first reaction does not occur, but the second does. (b) Suppose the reaction \(\mathrm{e}^{-}+\mathrm{e}^{+} \rightarrow 2 \gamma\) occurs and one of the photons travels toward Earth. What is the energy of the photon?
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