Some particle accelerators allow protons (p⁺) and antiprotons ...

Question

Some particle accelerators allow protons (p⁺) and antiprotons (p⁻) to circulate at equal speeds in opposite directions in a device called a storage ring. The particle beams cross each other at various points to cause p⁺ + p⁻ collisions. In one collision, the outcome is p⁺ + p⁻ → e⁺ + e⁻ + γ + γ, where γ represents a high-energy gamma-ray photon. The electron and positron are ejected from the collision at 0.9999995c and the gamma-ray photon wavelengths are found to be 1.0 x 10^-6 nm. What were the proton and antiproton speeds, as a fraction of c, prior to the collision?

Accepted Answer

Hey, everyone. So this problem is dealing with relativistic energy. Let's see what it's asking us. Envision a hypothetical scenario where a novel particle or a neutron moves to the right and its antiparticle or an anti utron moves to the left at the same speed. When these particles inevitably collide, they annihilate each other and produce two gamma ray photons of a wavelength 1.2 centimeters and a vitton and anti Vy pair with a speed of 0.92 multiplied by C or the speed of light. We're asked to determine the initial speed of the Routon and Anti Routon. In terms of sea before the collision, we're told that the mass of the neutron is 1.52 times 10 to the negative 27 kg. And the vitton is 8.21 times 10 to the negative 31 kg. So our multiple choice answers in terms of C are AC B 0.87 CC 0.57 C or D 0.64 C. So when we think about the conservation of energy, we have this novel particle, the Roon and that energy is equal to that of a photon that it produces plus that of the vitton that it produces. And so we could also say that to the energy of the Rron because you have the particle and the antiparticle and then you have two photons and then you have the particle and the antiparticle of the vitton. But we can also see that those twos cancel. And so we are left with that equation that the energy of the Rron is equal to the energy of that photon plus the energy of the Vyron. And so our relativistic energy equation which we'll use for our Vitton article, we can recall is equal to M multiplied by C squared multiplied by gamma where gamma is equal to, I'll write this out MT squared, gamma is equal to one divided by the square root of one minus V squared divided by C squared. And the energy of our photon is equal to H multiplied by C where age is planks constant C is our speed of light all divided by lambda. So we can solve for our vitton energy and our photon energy and then find that Roon energy and then use the relativistic energy equation to solve four E. So let's see what that looks like. We have E sub V is equal to. So the mass of the vitton is 8.21 times 10 to the negative 31 kg multiplied by C. That's our speed of light. We can recall as a constant three times 10 to the 8 m per second. That quantity squared all divided by the square root of one minus. And now the Von has a speed of 0.92 C. That quantity squared all divided by C squared, you plug that into our calculator and we get 1.89 times 10 to the negative 13 Joel. And now we need to find the energy from our photons. And so H or planks constant, we can recall is 6.63 times 10 to the 34 dual seconds multiplied by C again, that's three times 10 to the eighth meters per second, all divided by lambda, which is given to us in the problem as 1.2 fem time meters. So keeping everything in standard units, we can recall that a femto is 10 to the negative 15. And when we plug that into our calculator, we get 1.658 times 10 to the negative 10 jewels. And so the energy from the photon is obviously significant large, significantly larger than the energy from the von. And so when we add those two together, our energy of our retton is 1.66 times 10 to the negative 10 joules. And now we can use that same relativistic energy equation to solve for the speed of this particle. So I have our neutron energy is equal to M multiplied by C squared all divided by the square root of one minus V squared divided by C squared. And when we want to isolate that V squared term, we can, our first step will be the square root of one minus V squared divided by C squared is equal to MC square divided by the energy of a routon. And so the mass of a rut rut, excuse me was given as 1.52 times 10 to the negative 27 kg multiplied by C squared again, three times 10 to the eighth meters per second squared divided by the energy of the neutron, which we just solved for 1.66 times 10 to the negative 10 joules. And so what that gets us is the square root of one minus V squared divided by C squared is equal to 0.824. And then from there, we can square both sides and we'll add V squared divided by C squared to the right hand side. We subtract the 0.824 from the left hand side. And we'll have B squared is equal to 0.32 sea square. And then finally, we square, we take the square root of both sides and we have V is equal to 0.57 C. And so that is the final answer to this question. And when we look at our multiple choice answers, we can see that, that aligns with answer choice C so C is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

Key Concepts

Relativistic Speeds

Conservation of Energy and Momentum

Photon Wavelength and Energy Relation

Watch next