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Chapter 19: Problem 74

A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ -_{-1}^{0} e+_{+1}^{0} e \longrightarrow 2_{0}^{0} \gamma $$ Assuming that both g rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

Short Answer

Expert verified
The wavelength of the electromagnetic radiation produced during the annihilation of a positron and an electron can be calculated with the following formula: \(\lambda = \frac{hc}{E_{photon}}\), where \(h = 6.626 \times 10^{-34} Js\) is Planck's constant, \(c = 3.0 \times 10^8 m/s\) is the speed of light, and \(E_{photon}\) is the energy of a single photon. Substituting the given values, we get: \(\lambda = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{E_{photon}}\).

Step by step solution

01

Calculate the total energy produced

Using Einstein's formula, \( E = mc^2 \), for both the electron and the positron: \( E_{electron} = m_{electron}c^2 \) \( E_{positron} = m_{positron}c^2 \) Since the electron and positron have the same mass, the total energy produced will be: \( E_{total} = E_{electron} + E_{positron} = 2 m_e c^2 \) Where, \(m_e = 9.109 \times 10^{-31} kg \) is the mass of an electron/positron and \(c = 3.0 \times 10^8 m/s\) is the speed of light.
02

Calculate the energy of a single photon

Since two photons of equal energy are produced during the annihilation, the energy of a single photon is: \( E_{photon} = \frac{E_{total}}{2} \)
03

Calculate the wavelength (\(\lambda\)) of the electromagnetic radiation

Now we will use the energy-wavelength equation: \( E_{photon} = h\frac{c}{\lambda} \) Rearranging for the wavelength: \( \lambda = \frac{hc}{E_{photon}} \) Where, \(h = 6.626 \times 10^{-34} Js\) is the Planck's constant Now substituting the given values for the masses, speed of light, and Planck's constant, and the energy of a single photon that we calculated in step 2: \( \lambda = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{E_{photon}} \)
04

Calculate the result

Suppose \(m_e = 9.109 \times 10^{-31} kg\), \(c = 3.0 \times 10^8 m/s\), and \(h = 6.626 \times 10^{-34} Js\), then: \(E_{total}=2m_ec^2=2 \times (9.109 \times 10^{-31} kg) \times (3.0 \times 10^8 m/s)^2\) \(E_{photon}=\frac{E_{total}}{2}\) And finally, \( \lambda = \frac{(6.626 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{E_{photon}} \) This equation will give us the wavelength of the electromagnetic radiation produced during the annihilation of positron and electron.

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