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Chapter 18: Problem 2542

A gamma ray Photon creates an electron-Positron Pair. If the rest mass energy of an electron is \(0.5 \mathrm{MeV}\). and the total kinetic energy $0.7 \mathrm{MeV}$, then the energy of the gamma ray Photon must be (A) \(3.9 \mathrm{MeV}\) (B) \(1.78 \mathrm{MeV}\) (C) \(0.78 \mathrm{MeV}\) (D) \(0.16 \mathrm{MeV}\)

Short Answer

Expert verified
The energy of the gamma-ray photon is approximately \(1.7\,\mathrm{MeV}\), which is closest to option (B) \(1.78\,\mathrm{MeV}\).

Step by step solution

01

Determine the Rest Mass Energy of the Electron-Positron Pair

We are given that the rest mass energy of an electron is \(0.5\,\mathrm{MeV}\). Since an electron and a positron have the same rest mass energy, the total rest mass energy of the pair is: Total rest mass energy = Rest mass energy of electron + Rest mass energy of positron = \(0.5\,\mathrm{MeV} + 0.5\,\mathrm{MeV} = 1.0\,\mathrm{MeV}\)
02

Determine the Total Energy of the Electron-Positron Pair

Now, we need to find the total energy of the electron-positron pair including their kinetic energy. We are given that the total kinetic energy of the pair is \(0.7\,\mathrm{MeV}\). Therefore, the total energy of the electron-positron pair is: Total energy = Total rest mass energy + Total kinetic energy = \(1.0\,\mathrm{MeV} + 0.7\,\mathrm{MeV} = 1.7\,\mathrm{MeV}\)
03

Apply the conservation of energy

According to the principle of conservation of energy, the energy of the gamma-ray photon must be equal to the total energy of the electron-positron pair. Therefore, the energy of the gamma-ray photon is: Energy of gamma-ray photon = Total energy of electron-positron pair = \(1.7\,\mathrm{MeV}\)
04

Identify the correct option

Now that we have calculated the energy of the gamma-ray photon, we can compare it with the given options: (A) \(3.9\,\mathrm{MeV}\) (B) \(1.78\,\mathrm{MeV}\) (C) \(0.78\,\mathrm{MeV}\) (D) \(0.16\,\mathrm{MeV}\) Since the energy of the gamma-ray photon is \(1.7\,\mathrm{MeV}\), it is closest to option (B), which is \(1.78\,\mathrm{MeV}\). Hence, the correct answer is: (B) \(1.78\,\mathrm{MeV}\)

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Most popular questions from this chapter

In Rutherford experiment, the number of Particles scattered at \(90^{\circ}\) angle are 28 Per min. then the number of Particles at the angle \(120^{\circ}\) in Per min will be (A) 25 (B) \(12.0\) (C) 50 (D) 112

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It the radius of a nucleus of mass number 3 is \(\mathrm{R}\). then the radius of a nucleus of mass number 81 is (A) \(27 \mathrm{R}\) (B) \(9 \mathrm{R}\) (C) \(3 \mathrm{R}\) \((\mathrm{D})(27)^{(1 / 2)} \mathrm{R}\)

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