An electron traveling at 3.7×10⁵ m/s has an uncertainty in its velocity of 1.88×10⁵ m/s. What is the uncertainty in its position?

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hey everyone in this example we need to calculate our uncertainty and position of our particle and we want to recall that uncertainty in position is otherwise known as delta X. Were given information on the uncertainty of velocity, which we should recall is delta V in meters per second, which is actually the proper units. And were also given the mass of our particle in the proper units of kilograms. So we also want to recognize that our uncertainty and position should be in units of PICO meters. So our first step is to recall our formula for the Heisenberg principle of uncertainty and that begins with our uncertainty of position. Delta X multiplied by our massive our object. In this case is our particle. Then this is multiplied by our uncertainty and velocity delta V Set greater than or equal to Plank's constant divided by four times pi. So our next step is to plug in all of our knowns into this formula. So we don't know our uncertainty in position. That's what we're solving for. We multiply this by our given mass of our particle, which the problem tells us is this value here. So we'll plug that in 2.16 times 10 to the negative 28th power kilograms. And then this is multiplied by our uncertainty in velocity delta V which is given to us as 2.66 times 10 to the fifth power meters per second. This is then set greater than or equal to our numerator or in our numerator we have planks constant, Which we should recall is 6.626 times 10 to the negative 34th power jewels times seconds. And then in our denominator we still just have four times pi. So we can actually simplify this right hand side even further by recalling that our numerator plank's constant is equal to jewels time seconds, which can also be interpreted as kilograms times meters squared, divided by seconds squared. So we're going to substitute this in place of our jewels here. So what we should have in our next line is that our uncertainty of position multiplied by our uncertainty and velocity and then still multiplied by the mass of our object. Sorry, this is our uncertainty velocity 2.66 times 10 to the fifth power meters per second set greater than or equal to In our numerator we will have now 6.626 times 10 to the negative 34th power. And we're just substituting in for jules, kilograms times meters squared, divided by seconds squared. And then this is still multiplied by seconds as we saw here in this first step. So in our denominator we still just have four times pi and now we're just going to simplify both sides of our equation. So what we would have is that our uncertainty and position once we multiply our mass of our object and our uncertainty of velocity we should have the following value which is 5.70 for 56 times 10 to the negative 23rd power and we should have the units, kilograms times meters per second this is set greater than or equal to. We're going to take the entire calculation here in our calculators. But before we do so, let's go ahead and see how these units cancel out. So we'll be able to cancel out the seconds here because we have second squared in the denominator and then we have them here being multiplied in the numerator. So that would just leave us with one second unit. So when we multiply this entire quantity in our calculators, we can simplify so that we have 5.27- times 10 to the negative 35th power. And we should be left with kilograms times meters squared divided by just seconds in our denominator. And so to answer this question, we want to continue to isolate for delta X the uncertainty and position. So we're just going to simplify our left hand side by dividing both sides of our equation by the 5.7456 times 10 to the negative 23rd power, kilograms times meters per second term on both sides. And so this would fully cancel out this unit on the left hand side. And then on the right hand side we can go ahead and get rid of our kilogram unit as well as one of the meter in the numerator. And then the one in the denominator as well. And then we can also cancel out the second unit. So all we'll be left with is just one single meter unit in the numerator. And so what we should have when we simplify is that our uncertainty and position delta X is set greater than or equal to. We're gonna divide this quantity here in our calculators now that we've canceled out the units and that is going to give us a result equal to 9.177 times 10 to the negative 13th power meters. Because that is the only unit we're left with here in the numerator. And finally we want to recall that our delta X units are uncertainty in position should be in pickle meters. And so what we're going to do is basically go from meters to PICO meters. So we want to recall that our base unit meters gives us one peak a meter. And because PICO the prefix means 10 to the negative 12 power, we can plug that into our denominator so we can go ahead and now that we have this conversion factor outlined, we can cancel out meters. This leaves us with PICO meters as our final unit which is exactly what we want. And that is going to give us our uncertainty and position Delta X greater than or equal to a value of 0.9177 which we can actually just round 23 sig figs. Which is the minimum sig figs for this problem based on the information given To about 0.918 km. And so this here is going to complete this example as our final answer. So I hope that everything we reviewed was clear, but if you have any questions, please leave them down below, and I will see everyone in the next practice video.

An electron traveling at 3.7×10⁵ m/s has an uncertainty in its velocity of 1.88×10⁵ m/s. What is the uncertainty in its position?

Verified Solution

Key Concepts

Heisenberg Uncertainty Principle

Momentum

Calculating Uncertainty

An electron traveling at 3.7×105 m/s has an uncertainty in its velocity of 1.88×105 m/s. What is the uncertainty in its position?

Verified Solution

Key Concepts

Heisenberg Uncertainty Principle

Momentum

Calculating Uncertainty

An electron traveling at 3.7×10⁵ m/s has an uncertainty in its velocity of 1.88×10⁵ m/s. What is the uncertainty in its position?