(a) Find the radius of the 6^12 C nucleus. (b) Find the force of repulsion between a proton at

(a) Find the radius of the 6^12 C nucleus. (b) Find the...

(a) Find the radius of the ${ }_{6}^{12} \mathrm{C}$ nucleus. (b) Find the force of repulsion between a proton at the surface of a ${ }_{6}^{12} \mathrm{C}$ nucleus and the remaining five protons. (c) How much work (in MeV) has to be done to overcome this electric repulsion in transporting the last proton from a large distance up to the surface of the nucleus? (d) Repeat parts (a), (b), and (c) for ${ }_{92}^{238} \mathrm{U}$.

(a) Find the radius of the ${ }_{6}^{12} \mathrm{C}$ nucleus.
(b) Find the force of repulsion between a proton at the surface of a ${ }_{6}^{12} \mathrm{C}$ nucleus and the remaining five protons.
(c) How much work (in MeV) has to be done to overcome this electric repulsion in transporting the last proton from a large distance up to the surface of the nucleus?
(d) Repeat parts (a), (b), and (c) for ${ }_{92}^{238} \mathrm{U}$.

Key Concepts

Work and Energy Conversion in Nuclear Physics

Calculating the work involves determining the energy needed to overcome the electrostatic repulsion between protons. This energy is often calculated using the potential energy difference and then converted into common energy units like MeV for nuclear processes. This concept bridges classical mechanics with nuclear physics, highlighting how energy input relates to nuclear stability and reaction thresholds.

Coulomb’s Law

Coulomb’s law describes the electrostatic interaction between charged particles. In nuclear physics, it is used to calculate the repulsive force between protons within the nucleus. The law states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them. This is key in understanding how nucleons interact via electromagnetic repulsion.

Nuclear Radius Scaling

This concept refers to the empirical relationship that the size of a nucleus increases roughly as the cube root of its mass number (A). It is usually expressed as R = R? × A^(1/3), where R? is a constant. This relationship helps approximate the dimensions of different nuclei and is fundamental in nuclear structure and reaction studies.

Electrostatic Potential Energy

In a system of charges, the potential energy represents the work required to bring charges from infinity to a finite separation against the electrostatic force. In the context of nuclear physics, this concept helps determine the energy barrier (Coulomb barrier) that must be overcome to add a proton to a nucleus. It is crucial for analyzing nuclear reactions and processes such as fusion and radioactive decay.

Transcript

00:01 Now this question we are considering the decay of cobort 57 as well as carbon 40.

00:15 So for a decay to occur, it must be spontaneous.

00:20 That is, its cue value of the decay reaction must be positive.

00:28 If a decay would not occur if the q value is negative because it would mean that it needs energy input in order for the decay to occur, which is not possible.

00:42 Right for a decay process.

00:46 Now looking at the cobot 57 decay via positron decay, well, we can first write down the equation.

00:55 So for cobot to decay by a positron, the final product that we will have can be inferred via the man's number as well as the proton number.

01:16 So on the left, the total is 57, so on the right here must be 57.

01:21 Proton number is 27 total, so here must be 26.

01:26 This element is iron, so iron 57.

01:32 Now we can find the q value of this reaction, taking the change in the mass, multiplied by c square, which is the mass of cobot, 97, minus away.

01:52 So to do this, we will have to add 26, sorry, 27 electrons on both sides, in order to use the atomic mass.

02:05 Right, 27 electrons, then this will be a neutral atom, this will be a neutral atom plus an extra electron.

02:14 So we use the atomic masses of cobalt 57, minus away the atomic mass of iron, iron 57, minus away the mass of one electron and one positron.

02:33 So minus away two times the mass of an electron.

02:40 Multiplied by c square so the atomic masses can calculate can look up the table right for the 4 .2 we're going to use c square as 931 .5 m e v per u so the q value that we will get be negative 0 .187 mev just a negative negative value right and therefore since this is a negative value this reaction will not occur spontaneously and therefore this decay will not occur...

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