(a) The uncertainty in the y-component of a proton's position is 2.0x10^-12 m. What is the minimum uncertainty in a simultaneous measurement of the y-component of the proton's velocity?

Video transcript

Hey, everyone. Let's go through this practice problem. Consider an electron confined along the x direction in a quantum well of width 12 nanometers determine the minimum uncertainty of the velocity of the electron in the x direction. And we were given four options to choose from option A 4. multiplied by 10 to the power of three m per second. Option B 9.66 multiplied by 10 to the power of three m per second. Option C 1.16 multiplied by 10 to the power of five m per second. And option D 2.32 multiplied by 10 to the power of five m per second. We're looking for a relationship between an uncertainty in position and an uncertainty in velocity. So we can find a relationship between them using the Heisenberg uncertainty principle which states that the uncertainty in position of an object multiplied by the uncertainty in momentum is greater than or equal to the plank constant divided by four pi. And of course, recalled the plank constant as a value of 6. multiplied by 10 to the power of negative four jewel seconds. Now, the problem doesn't tell us the uncertainty in the electrons momentum. But we can figure that out anyway, if we recall that momentum is equal to a particle's mass multiplied by its velocity. So we can incorporate that into the plank equation by rewriting it as delta X the chain, the uncertainty in position multiplied by the mass of the electron multiplied by its uncertainty in velocity. And it should be noted that since we're looking for the minimum uncertainty, so the smallest possible value that the uncertainty of the velocity can have, that means that we're effectively looking for the value of the uncertainty where the left hand side of the uncertainty principle is exactly equal to the plank constant divided by four pi since this is where the uncertainty represents the smallest possible value of the uncertainty that is valid for the equation. So let's algebraically solve this equation for the uncertainty and the velocity by dividing both sides of the equation by delta X M. So we find an equation stating that the uncertainty in the velocity is equal to the plank constant divided by four pi multiplied by the mass of the electron multiplied by the uncertainty in its position. And of course, we're called the mass of an electron is equal to 9.11 multiplied by 10 of the power of negative kg. So all you have to do is plug in these numbers into a calculator. So the uncertainty and the velocity is equal to the plank constant 6.626, multiplied by 10th power of negative 34 dual seconds all divided by four pi 9. multiplied by 10. The power of negative 31 kg multiplied by the uncertainty in the position which is given to us as 12 in nanometers or in meters, 12 multiplied by 10 to the power of negative nine m. And if we put that into a calculator, then we find a value for the uncertainty of about 4. multiplied by 10, raised to the power of three m per second. And so that is the answer to the uncertainty of the velocity in this problem. And if we look at our options, we can see that this agrees with choice A 4.83 multiplied by 10 to the power of three m per second. So option A is our answer to this problem and that's all for now. I hope this video helped you out if you'd like more experience, please check out some of our other videos which will give you more practice with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.