The nuclei of large atoms, such as uranium, with 92 protons, can ...

Question

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately 7.4×10−15 m. (b) What magnitude of electric field does it produce at the distance of the electrons, which is about 1.0×10−10 m?

Accepted Answer

Welcome back everybody. We are taking a look at a helium atom. Now this helium atom is going to have a spherical nucleus containing two protons and it is surrounded by two electrons that are orbiting that nucleus. Now we are told that the radius of orbit is one times 10 to the negative 12 m. And we need to find the strength of the electric field generated by the nucleus at the electrons orbit or this. We're going to use God's law and we're going to use two of his formulas here. One of them states that the electric flux is equal to the electric field times the area. Right? Which in this case, since we're dealing with this sphere, we're going to have the electric field times the surface area, which is four pi R squared. Now we also know from Galaxies law that the electric flux is equal to the Q enclosed, divided by the electric purgative itty constant, which is just epsilon. Not I'm going to combine these two equations together and we get that E times four pi R squared is equal to Q enclosed, divided by epsilon. Not now, I'm gonna divide both sides. 54 I R squared. This will cancel out on the left and we are left with this term right here. Now, I'm actually gonna separate this into two fractions and you'll see why in just a sec when we start plugging in numbers, I'm gonna separate this into 1/4 pi times epsilon, not times our Q enclosed. All over our radius of orbit squared. Now that we have our formula for our electric field. Let's go ahead and plug in some values. So what is this term? Right here, 1/4 pi times the electric relativity constant. Well, that is just cool. Um is constant, which is just 8.99 times 10 to the ninth. This of course is going to be multiplied by the charging closed. Well, the charging closes. We have two spherical protons, so it'll be two times the charge of a proton, which is 1.6 times 10 to the negative 19th. All divided by r radius of orbit, which is 10 to the negative 12 squared. And when you plug all of this in your calculator, you get that the strength of the electric field generated by the nucleus at the orbit is 2.9 times 10 to the 15th. Newtons per um which is equivalent to our answer choice of D. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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