The last step in the carbon cycle CNO-I that takes place...

The last step in the carbon cycle CNO-I that takes place inside stars is $\mathrm{p}+{ }^{15} \mathrm{N} \rightarrow{ }^{12} \mathrm{C}+(?) .$ (a) Show that the reaction product "(?)" must be an alpha particle. (b) How much energy is released by this step of the cycle? (c) In order for this reaction to occur, the proton must come into contact with the nitrogen nucleus. Calculate the distance $d$ between their centers when they just "touch." (d) If the proton and nitrogen nucleus are initially far apart, what is the minimum value of their total kinetic energy necessary to bring the two into contact?

Key Concepts

Conservation Laws in Nuclear Reactions

In nuclear reactions, the conservation of mass number (total nucleons) and charge is paramount. The particles entering the reaction must produce particles whose combined mass numbers and charges equal those of the reactants. In the given reaction, comparing the nucleon count and charge on both sides leads to the determination that the unidentified product must be an alpha particle. This analysis is a cornerstone of understanding nuclear reaction pathways in astrophysics and nuclear physics.

Binding Energy and Q-value

The energy released in a nuclear reaction, commonly known as the Q-value, arises from the difference in binding energy between the reactants and products. Binding energy reflects how tightly nucleons are held together within a nucleus, and a reaction that produces a more stable configuration releases energy. In this context, calculating the mass defect and converting it to energy using Einstein’s mass-energy equivalence (E = mc^2) gives the energy released in the final step of the carbon cycle.

Coulomb Barrier

The Coulomb barrier is the energy barrier due to electrostatic repulsion that charged particles must overcome to get close enough for the strong nuclear force to bind them together. In the reaction described, a proton must approach a nitrogen nucleus, and the repulsion between their positively charged protons creates this barrier. Understanding the Coulomb barrier is essential for calculating the minimum kinetic energy required for the reaction to occur, as it dictates the need for sufficiently high velocities in stellar environments.

Nuclear Contact Distance

The nuclear contact distance is determined by the radii of the interacting nuclei and represents the distance at which the surfaces of two nuclei are in contact, facilitating the nuclear reaction. Typically, nuclear radii are estimated using the formula R = r0A^(1/3), where r0 is a constant and A is the mass number of the nucleus. This concept helps in estimating the distance d between the centers of the proton and nitrogen nucleus at the point where their surfaces just touch, which is crucial for understanding the onset of strong nuclear interaction after overcoming the Coulomb barrier.

Transcript

00:01 Hello, and in this question here, we're going to be looking at the final stage of the cno cycle.

00:08 So we have a proton, and there's a nuclear reaction with this proton and a nitrogen 15 atom, and it goes into a carbon 12 atom plus some unknown particle.

00:22 So we want to find out what this unknown particle x is.

00:26 Well, to do this, before the reaction, we note that we have 8 .5.

00:31 Protons okay eight protons one of them comes from the single proton here and we have seven protons from the nitrogen atom we also have eight neutrons and these eight neutrons are entirely from the nitrogen 15 atom so the bottom number this seven represents the number of protons in the nucleus and the top number of 15 represents the number of nucleons so 15 minus seven gives eight neutrons this becomes, on the other side, what carbon 6, carbon 12, has 6 protons and 6 neutrons.

01:13 Okay? so the 6 protons from the bottom number and 12 minus 6 gives 6 neutrons.

01:23 So this means we must have plus we're going to have a protons from particle x and we're going to have b neutrons from particle x.

01:34 Particle x as well.

01:37 So by having a conservation of the number of protons and neutrons, we can work out that a must be equal to 2 and b must be equal to 2.

01:47 So this particle x contains two protons plus two neutrons and this is what an alpha particle.

01:58 Okay? so x is equal to alpha or alpha particle.

02:10 So the next part of the question is asking us to determine the energy released in this collision or in this reaction.

02:23 So the energy will be the change in mass of the initial state and the final state multiplied by c squared is equal to e.

02:36 So the change in mass is converted into energy from einstein's m equals c e equals nc squared equation.

02:44 So we need to calculate this change in mass.

02:47 Well, the mass of the initial state is equal to the mass of the proton plus the mass of the nitrogen 15 atom minus the mass of the carbon 12 atom minus the mass of the alpha particle.

03:04 And if we multiply by this by c squared, we'll get the energy released in this reaction.

03:09 Well, from the appendix in the book, we know that the mass of the proton is equal to 1 .007825u.

03:25 The mass of the nitrogen -15 atom is equal to 15 .0001 -089 u.

03:39 The mass of the carbon 12 atom, well this is the very definition of what u is.

03:47 So this is just equal to 12 times u as u is defined as 1 .12 of the mass of the carbon 12 atom.

03:56 And the mass of the alpha particle is equal to 4 .0026033u.

04:05 And if you're struggling to find this, this is the mass of the alpha particle, is equal to the mass of the helium nuclei with four nucleons in it.

04:15 Okay? so just using the fact that u is equal to 1 .66 times 10 to the minus 27 kg.

04:27 We can calculate what this delta m is, as we know all of these masses...

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