Question

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

Short answer

The total rest energy before annihilation can be calculated as \(E_0 = 2m_e c^2\), where \(m_e = 9.109 \times 10^{-31} kg\) and \(c = 3 \times 10^8 m/s\). The energy of a single photon after annihilation is \(\frac{E_0}{2}\). Using the energy of a photon equation, \(E_{\gamma\_single} = hf_{\gamma}\), we can find the frequency of the electromagnetic radiation, where \(h = 6.63 \times 10^{-34} Js\). Finally, we can calculate the wavelength using the speed of light equation, \(c = \lambda f_{\gamma}\), which results in the wavelength \(\lambda = \frac{c}{f_{\gamma}}\).

Step-by-step solution

Calculate the total energy before annihilation

First, we need to find the total energy before the annihilation. This consists of the rest energy of the electron (e-) and the positron (e+). The rest mass of electron m_e = 9.109 × 10^(-31) kg The rest mass of positron m_p = m_e (they have the same mass) The speed of light c = 3 × 10^8 m/s The total rest energy E_0 = m_e c^2 + m_p c^2 = 2m_e c^2

Calculate the energy of the photons after annihilation

The conservation of energy principle states that the total energy before annihilation must equal the total energy after annihilation. Therefore, the energy of the two photons (E_gamma) produced in this reaction is equal to the total rest energy E_0, which we calculated in the previous step. E_gamma = E_0 Since both photons have the same energy, we can calculate the energy for a single photon: E_gamma_single = E_gamma / 2

Find the frequency of electromagnetic radiation

Now, we'll use the energy of a photon equation to find the frequency of the electromagnetic radiation: E_gamma_single = h * f_gamma where h is Planck's constant, 6.63 × 10^(-34) Js Solving for the frequency (f_gamma): f_gamma = E_gamma_single / h

Calculate the wavelength of the electromagnetic radiation

Finally, we'll use the speed of light equation to find the wavelength of the electromagnetic radiation produced: c = λ * f_gamma Solving for the wavelength (λ): λ = c / f_gamma Now we can plug in the numbers and calculate the wavelength of the electromagnetic radiation produced.

Key concepts

Positron-Electron Annihilation

When a positron, the antimatter counterpart of an electron, collides with an electron, they undergo a process known as positron-electron annihilation.

During this event, their masses are not simply lost but are converted into energy in the form of gamma-ray photons. It's fascinating to imagine that these particles essentially vanish and in their place, pure energy is emitted. This conversion adheres to the principles of the conservation of energy and momentum.

• The process is represented by the equation: \( e^+ + e^- \rightarrow 2\gamma \)
• Here, \( e^+ \) and \( e^- \) are the positron and electron, respectively, and \( \gamma \) represents the gamma-ray photons produced.
• This reaction is instrumental in applications like PET scans in medical imaging, where positron-emitting substances are used to detect metabolic processes in the body.

Energy-Mass Equivalence

The concept of energy-mass equivalence was made famous by Albert Einstein's equation \( E = mc^2 \).

This equation signifies that energy (E) and mass (m) are interchangeable; they are different forms of the same thing. In the realm of particle physics.

• When a particle and its antiparticle annihilate, the total rest mass is converted into energy.
• For a positron-electron annihilation, the mass of both the positron and electron (\( 9.109 \times 10^{-31} \) kg) is converted into energy.
• The speed of light (\( c \)) in this equation is a constant (\( 3 \times 10^8 \) m/s), representing the speed at which light travels in a vacuum.
• Thus, the energy released during annihilation is given by \( E_0 = 2m_e c^2 \) which is the sum of the rest energies of both particles.

Photon Energy Calculation

The final stage of understanding the positron-electron annihilation involves calculating the photon energy and determining the wavelength of the electromagnetic radiation produced.

This can be achieved through a series of steps where concepts of quantum mechanics come into play, specifically the relationship between the energy of a photon and its frequency (\( E = hf \)).

• Planck's constant (\( h \)), a fundamental constant of nature (\( 6.63 \times 10^{-34} \) Js), is critical for this calculation.
• By dividing the total energy by Planck's constant, the frequency of the photon can be found (\( f = \frac{E}{h} \) ).
• The speed of light (\( c \) ) also relates the photon's frequency to its wavelength (\( c = \lambda f \)).
• By rearranging this equation, we can solve for the wavelength (\( \lambda = \frac{c}{f} \)), thus finding the length of the electromagnetic waves resulting from the annihilation event.