A gamma ray (a high-energy photon of light) can produce a particle- antiparticle pair, each particle with mass 8.68 MeV/c2 when it enters the electric field of a heavy nucleus: γ ! λ+ + λ− .
What minimum γ-ray energy is required to accomplish this task?
I have searched but can't come up with a numerical answer in MeV units. Please Help
You know the mass created: 2*8.68Mev/c^2
so energy= masscreated*c^2=2*8.68Mev
thank you very, i did the calculations and it was simple math after your explanation..
i got answer as 17.36 MeV
=]
To determine the minimum gamma-ray energy required to produce a particle-antiparticle pair, we can use the principle of conservation of energy and momentum. The process involves the creation of a particle (lambda plus, λ+) with its antiparticle (lambda minus, λ-) when a gamma ray (γ) interacts with the electric field of a heavy nucleus.
Since the initial and final states of the system need to satisfy the conservation laws, we need to consider the following:
1. Conservation of energy: The energy of the gamma ray must be equal to the combined energy of the particle and antiparticle produced.
2. Conservation of momentum: The momentum of the gamma ray must be equal to the combined momentum of the particle and antiparticle produced.
Now, let's calculate the minimum gamma-ray energy required:
1. Conservation of energy:
The energy of the gamma ray (γ) can be calculated using the equation E = mc^2, where E is the energy, m is the mass, and c is the speed of light. Since the particle-antiparticle pair has a combined mass of 8.68 MeV/c^2, we can express the required energy as E = 2 * 8.68 MeV. Therefore, the minimum energy of the gamma ray is 2 * 8.68 MeV.
2. Conservation of momentum:
The momentum (p) can be calculated using the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. Since the gamma ray interacts with the electric field of a heavy nucleus, it will be moving at the speed of light (c), and its momentum is given by p = E/c, where E is the energy. Therefore, the momentum of the gamma ray is (2 * 8.68 MeV/c) / c = 2 * 8.68 MeV/c.
Thus, the minimum gamma-ray energy required to produce the particle-antiparticle pair is 2 * 8.68 MeV, or 17.36 MeV.
To determine the minimum gamma-ray energy required to produce a particle-antiparticle pair, we need to consider the conservation of energy and momentum.
The basic idea is that the minimum energy of the gamma-ray needs to be at least equal to the total rest mass energy of the produced particles.
Let's assume the minimum gamma-ray energy required is Eγ. This energy is used to create a particle and an antiparticle, each with mass 8.68 MeV/c^2. Therefore, the total rest mass energy (E_rest) of the produced particles is equal to 2 times the mass times the speed of light squared (2mc^2):
E_rest = 2 * (8.68 MeV/c^2) * (c^2)
Where c is the speed of light. The mass-energy equivalence principle states that E = mc^2, where m is the particle's mass and c^2 is the conversion factor between mass and energy.
Now, we can substitute the known values into the equation:
E_rest = 2 * (8.68 MeV/c^2) * (c^2)
= 2 * (8.68 MeV)
= 17.36 MeV
Hence, the minimum gamma-ray energy required to produce a particle-antiparticle pair with each particle having a mass of 8.68 MeV/c^2 is 17.36 MeV.