(a) If you treat an electron as a classical spherical object with...

Question

(a) If you treat an electron as a classical spherical object with a radius of 1.0 * 10-17 m, what angular speed is necessary to produce a spin angular momentum of magnitude 23 4 U? (b) Use v = rv and the result of part (a) to calculate the speed v of a point at the electron's equator. What does your result suggest about the validity of this model?

Accepted Answer

Hey everyone. So this problem is dealing with inertia at an atomic level. Let's see what it's asking us consider a negatively charged particle, which is an electron that is assumed to be perfectly spherical in shape, the negatively charged elementary particle that orbits the atom's nuclei has a radius of 2.14 times 10 to the negative 14 m. And the magnitude of its spin angular momentum is the square root of 6/7 H bar find its angular speed given by omega and the speed V of a point at the equator of the negatively charged particle. What can we say about the model's validity based on these results? So our multiple choice answers are given here. And the key to solving this equation is to recall that the moment of inertia is given by I equals 2/5 M R squared for a sphere. And in terms of angular momentum in angular velocity, I equals L which is the angular momentum divided by omega which is the angular velocity. And so setting these two equations equal to each other and isolating omega which we are solving for, we get five L divided by two M R squared where we now have everything we need to plug in and solve for our omega. So we have five multiplied by our angular momentum which was given as the squared of 67 multiplied by H bar. We can recall H bar planks reduce constant variable or planes reduced constant is equal to 1.55 times 10 to the negative 34 dual seconds. All of that divided by two multiplied by the mass of an electron. We can recall is 9.11 times 10 to the negative kg multiplied by the radius 2.14 times 10 to the negative 14 m. And that quantity squared and that was given to us in the problem. And so when we plug that into our calculator, we get a angular velocity of 5.85 times 10 to the 23 radiant per second. And so we can now use that angular velocity to solve or speed. We recall that the equation for speed in terms of angular velocity is simply R multiplied by omega. So again, the radius given to us in the problem is 2.14 times 10 to the 14 m is at the equator or at the um maximum value for the radius multiplied by omega which we just solve for 5.85 times 10 to the radiant per second. We get a speed of 1.25 times to the 10 m per second. And so the question also asks about the validity of the model. So since our speed is two orders of magnitude higher than the speed of light, you can recall the speed of light is times 10 to the eighth meters per second, which is the maximum speed for a matter um or energy because this speed was calculated as two orders of magnitude higher, the model is invalid because it's just not physically possible. And so all of that aligns with answer choice B so B is the correct answer for this problem. So all we have for this one, we'll see you in the next video.

Watch next