An electron is to be accelerated from 3.00x10^6 m/s to 8.00x10^6 m/s. (a) Through what potential difference must the electron pass to accomplish this?

Question

Accepted Answer

Hey, everyone in this problem, we have a proton that has a speed of 3.5 times 10 to the 6 m per second. The speed of the proton should be increased to 7.5 times 10 to the 6 m per second. And we're asked to determine the potential difference along the protons path that will achieve that increase in speed. OK. All right. So we're talking about potential difference. OK. We're talking about delta V big V. OK. And that's what we're looking for. What is that change in potential, right? All right. Now we're thinking about potential. We're trying to figure out what kind of equations we have to relate to potential. OK. We're given speeds, we're given potential. The potential is gonna relate to the potential energy. Our speeds relate to our kinetic energy. Well, let's use our conservation of mechanical energy. OK. We have no net external forces. So we have the conservation of mechanical energy which tells us that the initial kinetic energy cannot plus the initial potential energy U not, is equal to the final kinetic energy KF plus the final potential energy. UF OK. So the initial mechanical energy equals the final mechanical energy. Now, let's recall what is kinetic energy? Kinetic energy is one half MV squared. OK. Where V is the speed? And then we're talking little V so we can write this as one half M little V knot squared. OK. And on the right hand side be at one half M little VF squared. OK. And then what about the potential energy? Big U? OK. We're talking about protons, the potential energy is just gonna be QVK or Q is the charge and V is the potential. And so we get Q knot, OK? And V knot and this is gonna be big V knot. OK? Don't confuse the little V for speed with the big V for potential. OK? And similarly, on the right hand side, we get QF big VF OK? All right. So we have an equation that has the potential V nought and VF OK. That are going to be part of our potential difference. Delta V. Let's try to rearrange this OK. I'm just gonna write up here the change in voltage or sorry, the change in potential delta V is going to be big VF minus V naught. OK. The final potential minus the initial potential. So let's try to rearrange this equation to have VF minus V naught. So, so we're gonna move one half MVF squared to the left hand side. Here we go one half M times little V not squared minus little VF squared. And on the right hand side, we're gonna move this Q knot, V knot over. We're gonna get Q, OK. Now, Q knot and QF, well, this is a proton. OK. So before with the initial speed, it's a proton with the, the final speed it's still a proton. So it has the same charge is a Q knot is going to equal QF, OK? And we'll just call both of them Q. OK? So this is just gonna be Q times big VF minus big V naught. OK. Well, what do you see here? VF minus V naught. That's a potential difference. OK. That change in potential. So let's isolate that VF minus V nought. OK? We're gonna divide by Q, we have M times little V knot squared minus little VF squared divided by two Q. And again, this is delta V. So this is what we're looking for. Let's go ahead and substitute in the numbers that we have. OK? The mass, this is gonna be the mass of a proton. OK. This you can look up in a table in your textbook or that your professor has provided, this is gonna be 1.67 times 10 to the negative 27 kg. Now the initial speed V knot we're told is 3.5 times 10 to the 6 m per second. So we get 3.5 times 10 to the 6 m per second all squared. And then we're subtracting the final speed which is 7.5 times 10 to the 6 m per 2nd, 7.5 times 10 to the 6 m per second all squared. And we're gonna divide by two Q. OK? And the charge of a proton Q is 1.6 times 10 to the negative 19 Coomes. OK. And again, that's something you can look up in a table in your textbook or that your professor has provided. All right, let's give ourselves some more room to work. We're gonna have delta V is equal to 1.67 times 10 to the negative 27 kg. OK. Simplifying the change in speeds here with the square, we're gonna get negative 4.4 times 10 to the 13 and the unit here is meters squared per second squared and watch the sun. We have a negative here today because the speed increases and we have V minus VF and this is gonna be divided by 3.2 times 10 to the negative 19 columns. All right, this gives a delta V of negative 2.29 625. In our unit, we have kilogram meter squared per second square. That's a jewel and sorry, I missed. This is times 10 to the five. OK. All right. So again, kilogram meter squared per second squared is a jewel. So we get jewel per Coolum. OK. The unit we want and if we approximate, we get negative 2.3 times 10 to the five jeel per coon. That's a volt. OK. And so the potential difference that we want in order to have that change in speed is negative 2.3 times 10 to the five volts. If we go up to our answer choices, we see that that is going to be answer choice. D thanks everyone for watching. I hope this video helped see you in the next one.

An electron is to be accelerated from 3.00x10^6 m/s to 8.00x10^6 m/s. (a) Through what potential difference must the electron pass to accomplish this?

Verified Solution

Key Concepts

Kinetic Energy

Electric Potential Difference

Conservation of Energy