A proton in a linear accelerator has a de Broglie wavelength of 132 pm. What is the speed of the proton?

Video transcript

Hey everyone in this example, we need to find the speed of an electron with the D, broccoli wavelength of 162 PK meters. So what we want to first recognize is that because we want to find the speed of an electron, we're going to be looking for velocity. And so we want to recall our formula which states that our wavelength is equal to Planck's constant, divided by the mass of our object in this case our electron and then multiplied by velocity. And we're going to be isolating for V so that we can get our answer. So we want to make note of our mass of our electron for the denominator. And when we look that up, we should recall that it's equal to 9.11 times 10 to the negative 31st power kilograms. So our next step we should recognize is that are given wavelength is in units of PICO meters. And we want to convert that to meters. So we'll do that conversion here below. We have 162 PK meters and we want to multiply by the following conversion factor where we place PICO in our PICO meter in our denominator and meters in our numerator. And we should recall that our prefix PICO tells us that we have 10 to the negative 12 power of our base unit meters. So we can go ahead and cancel out PICO meters leaving us with meters. And this is going to give us our proper wavelength as 1.62 times 10 to the negative 10th power meters. So now we can go ahead and begin our formula. So on our left hand side we should have our wavelength that we just found above. And then we're setting that equal to our numerator, where we should recall plank's constant, which is 6.626 times 10 to the negative 34th power jewels, times seconds. And then in the denominator we're plugging in that mass of electron, which we said is 9.11 times 10 to the negative 31st power kilograms. And then multiplied by velocity, which is what we're isolating for. Now. Before we go ahead and isolate for velocity, we want to recall that um jules can be interpreted also as kilograms times meters squared, divided by seconds squared. So we're going to substitute this as a conversion factor in place of jewels, in our planks constant. And so in our next line we're just going to apply that substitution. And so our numerator should now look like 6.626 times 10 to the negative 34th power kilograms times meters squared, divided by seconds squared. And then we're still multiplying by the seconds that was present there originally in our numerator. And so in our denominator we want to keep that the same. So we have 9.11 times 10 to the negative 31st power kilograms times velocity. And at this point we can go ahead and actually isolate for velocity by multiplying by V on both sides of our equation. So this allows us to cancel out that variable on the right hand side. And so now what we'll have is that we have 1.62 times 10 to the negative 10th power times velocity set equal to our numerator of plank's constant, still multiplied by seconds. And then in our denominator we still have our massive R electron 9.11 times 10 to the negative 31st power kilograms. So we want to go ahead and continue to completely isolate velocity. So we're going to go ahead and divide both sides by our wavelength, which is 1.62 times 10 to the negative 10th power. And we're doing so on both sides. So we can actually go ahead and multiply that here in our denominator and we should still have our units of meters for our wavelength present in our formula at this point. So to make things simple, let's go ahead and do our calculation for this step on the right hand side so that we can come up with our value for our final line in this solution for velocity. And so the way the units are going to cancel out is we're gonna be able to get rid of kilograms because in the numerator appear. And then in the denominator down here we can get rid of one of our seconds here and then one of them here, so that we're left with just seconds to the first power. And then in our numerator, we can also cancel out one of the meters with the one down here so that we're left with meters to the first power. And so now we can go ahead and type the calculation for our quotient here and our calculators. So what that's going to give us is our velocity of our electron equal to a value of 4.48 nine times 10 to the sixth power. And then our units, we were just left with meters to the first divided by seconds to the first. So that's just meters per second. And we can go ahead and round this out 2366. So that our final answer is 4.49 times 10 to the sixth power meters per second. And so our official answer to this question is going to be that our speed of our electron or velocity is equal to 4.49 times 10 to the six power meters per second. And this is going to correspond to choice a in our multiple choice set. So this will complete this example. If you have any questions, please leave them down below and I will see everyone in the next practice video