The energy required to construct a uniformly charged sphere...

The energy required to construct a uniformly charged sphere of total charge $Q$ and radius $R$ is $U=3 k, Q^{2} / 5 R,$ where $k,$ is the Coulomb constant (see Problem 64 ). Assume a 40 . Ca nucleus contains 20 protons uniformly distributed in a spherical volume. (a) How much energy is required to counter their electrical repulsion according to the above equation? Suggestion: First calculate the radius of a "Ca nucleus. (b) Calculate the binding energy of $? 0$ Ca. (c) Explain what you can conclude from comparing the result of part (b) with that of part (a).

Key Concepts

Coulomb Energy of a Uniformly Charged Sphere

This concept explains the method of calculating the electrostatic energy stored in a continuous, uniformly charged spherical distribution. The derivation involves integrating Coulomb's law over the sphere's volume, resulting in a formula that quantifies the work needed to assemble the sphere by countering the electric repulsion among its charge elements.

Nuclear Radius Scaling

Nuclear radius scaling involves the empirical relationship that relates the size of a nucleus to its mass number, typically expressed as R = r?A^(1/3). This concept is essential for estimating the spatial dimensions of nuclei and is crucial when applying formulas that depend on the radius, such as the calculation of internal electrostatic energies.

Nuclear Binding Energy

Nuclear binding energy is the energy required to disassemble a nucleus into its individual nucleons. It represents a measure of nuclear stability, where a higher binding energy indicates a more stable nucleus. The concept is rooted in the mass defect and Einstein’s mass-energy equivalence principle, relating the binding energy to the difference in mass between the nucleus and its constituent protons and neutrons.

Electrostatic Repulsion in Nuclei

This concept addresses the repulsive forces between protons within the nucleus due to their positive charges. It involves understanding how the energy associated with overcoming these repulsive forces can be calculated using electrostatics, and how this repulsion is counterbalanced by the attractive nuclear forces, contributing to the overall stability and structure of the nucleus.

Transcript

00:01 In this question we are given the kulum potential energy.

00:06 This is given as 3 ke times q square over 5r.

00:20 What we want to find is to find what is the this potential energy for our calcium 40 nucleus.

00:32 This will give us what the electrostatic energy potential energy that is required to be overcome in order to keep the nuclei together right so basically this is like the repulsion energy all right so we have the charge of the calcium right so calcium basically we have 20 protons right 20 protons and so q is just 20 times the elementary charge.

01:15 We know the ke, this is the constant 1 over 4 pi epsilon knot, which is equals to 8 .99 times 10 to power 9, it's just a constant.

01:28 R, we need to find why is the radius.

01:30 So radius for our calcium folly, we can use the formula, r0 times a to power 1 3rd, where r0 is a constant 1 .2 times 10 power minus 15 meters and a is the nucleon number where in this case is just 40 right we have total number 40 number of nuclei new clons sorry so we take 1 .2 times 10 power minus 15 times 40 to the power of 1 third and we should get the radii to be 4 .1 times 10 power minus 15 and go going to use this radius for the equation.

02:27 So now we can finally look at what is the potential energy.

02:31 In this case, 3 times q, q square.

02:48 So we need to square this divided by 5 times the radius.

03:00 This will give us in terms of si units which is 1 .35 times 10 .10 .10 .10 .10 .1 .12.

03:10 It's a pretty small number so usually we want to come into 2.

03:13 M .ev or ev can divide by 1 .6 10 power minus 13 to convert from juice to mep...

Please give Ace some feedback