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Chapter 39: Problem 40

Suppose a neutral pion at rest decays into two identical photons. a) What is the energy of each photon? b) What is the frequency of each photon? c) To what part of the electromagnetic spectrum does this correspond?

Short Answer

Expert verified
Answer: The energy of each photon is 67.5 MeV, the frequency is 1.63 × 10^(23) Hz, and they belong to the gamma rays region of the electromagnetic spectrum.

Step by step solution

01

Calculate the energy of the pion

Using the mass-energy equivalence equation, we need to first find the energy of the pion before decay. The mass of a neutral pion is approximately 135 MeV/c², where MeV is million electron volts and c is the speed of light. Therefore, the energy of the pion is: E_pion = (mass of pion)* c^2 E_pion = 135 MeV Now, we know that the pion will decay into two identical photons. Thus, as per conservation of energy, the total energy of the two photons should be equal to the energy of the pion. Energy of two photons = E_pion = 135 MeV
02

Calculate energy of each photon

Since the two photons are identical, they will have equal energy. Thus, we can calculate the energy of each photon as follows: Energy of each photon = (Energy of two photons) / 2 Energy of each photon = (135 MeV) / 2 Energy of each photon = 67.5 MeV
03

Calculate frequency of each photon

We will use the energy of a photon equation to determine the frequency of each photon: E_photon = h * f Where, E_photon = 67.5 MeV and h (Planck's constant) = 4.135667696 × 10^(-15) eV*s. Solving for f: f = E_photon / h f = (67.5 × 10^6 eV) / (4.135667696 × 10^(-15) eV*s) f ≈ 1.63 × 10^(23) Hz
04

Determine the corresponding part of the electromagnetic spectrum

To determine which part of the electromagnetic spectrum these photons belong to, we can compare their frequency to the following ranges: - Radio waves: < 10^9 Hz - Microwaves: 10^9 to 10^12 Hz - Infrared: 10^12 to 10^14 Hz - Visible light: 10^14 to 10^15 Hz - Ultraviolet: 10^15 to 10^17 Hz - X-rays: 10^17 to 10^20 Hz - Gamma rays: > 10^20 Hz Since the frequency of our photons (1.63 × 10^(23) Hz) is higher than 10^20 Hz, they belong to the gamma rays part of the electromagnetic spectrum. To summarize: a) The energy of each photon is 67.5 MeV. b) The frequency of each photon is 1.63 × 10^(23) Hz. c) The photons correspond to the gamma rays region of the electromagnetic spectrum.

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