The last step in the carbon cycle that takes place inside...

The last step in the carbon cycle that takes place inside stars is $\mathrm{p}+{ }^{15} \mathrm{~N} \rightarrow{ }^{12} \mathrm{C}+(?) .$ This step releases 5.00 MeV of energy. (a) Show that the reaction product "(?)" must be an alpha particle. (b) Calculate the atomic mass of helium-4 from the information given. (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?

The last step in the carbon cycle that takes place inside stars is $\mathrm{p}+{ }^{15} \mathrm{~N} \rightarrow{ }^{12} \mathrm{C}+(?) .$ This step releases
5.00 MeV of energy. (a) Show that the reaction product "(?)" must be an alpha particle. (b) Calculate the atomic mass of helium-4 from the information given.
(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

Nuclear Radii and the Coulomb Barrier

The distance at which two nuclei 'touch' is derived from the sum of their nuclear radii, typically estimated using a formula proportional to the cube root of the mass number. This concept is essential for evaluating the Coulomb repulsion (the electrostatic force between positively charged nuclei) and calculating the kinetic energy required to overcome this barrier for a reaction to occur.

Atomic Mass Unit Conversions

Understanding the conversion between atomic mass units and energy (using the relation that 1 amu corresponds to about 931.5 MeV) is crucial for linking observed energy releases in nuclear reactions to minute changes in mass, thereby allowing the determination of nuclear masses.

Stellar Nucleosynthesis

This concept involves the processes by which elements are formed within stars through nuclear reactions. In the context of the carbon cycle, it addresses transformations among lighter nuclei into heavier ones, demonstrating how specific fusion reactions, like the one given, contribute to the buildup of elements in stellar interiors.

Conservation Laws in Nuclear Reactions

Conservation of energy and nucleon number are key when determining reaction products. By applying these laws, one can deduce the identity of unknown reaction fragments—in this case, confirming that the unspecified product must be an alpha particle to conserve both mass and charge.

Mass-Energy Equivalence and Nuclear Binding Energy

This principle (E=mc^2) connects the mass defect in a nuclear reaction to the energy released. In nuclear reactions, slight differences in mass between reactants and products manifest as significant energy releases, enabling calculations such as determining the atomic mass of a nucleus from the measured energy output.

Transcript

00:01 In this question we are given this particular nuclear reaction of proton with nitrogen 15, producing carbon with an unknown x, all right unknown project x.

00:15 And to figure out what x is, we look at the conservation of the nucleon number as well as the proton number.

00:23 So they both must be conserved from the left -hand side to the right -hand side.

00:28 Now looking at the nucleon number first, on the left side we have 1 plus 15, there is 16 total of a nuclear number.

00:37 So on the right we must have 12 plus something to give us 16.

00:41 That's 4.

00:42 So x must have a nucleon number of 4.

00:45 For the proton number, we have on the left hand side total of 8, 1 plus 7.

00:51 So on the right, the proton number for x must be 2.

00:56 And looking up the periodic table for an element with 2 protons, that is helium.

01:04 And so our unknown product over here must be helium 4.

01:09 And this is actually alpha particle.

01:15 Now next, we want to find out what is the atomic mass of this helium 4 from the information.

01:22 And the most important information that they have given us is that the energy released is actually 5mv for this reaction.

01:32 Now what does this tell us? this tells us the change in the mass from the products and the reactants, the difference in the mass, multiplied by c squared.

01:46 And this is very important, this change in the mass.

01:50 Because the change in the mass is actually the mass of carbon, plus the mass of our alpha particle, subtracted by the mass of the protons, plus the mass of the mass of the protons, plus the mass of the carbon, of the nitrogen and this is what we want to know why we assume that we have all of these other values which is the carbon the proton and the nitrogen right and this carbon is 12 .00 for the proton 0 .007 825 and for the nitrogen it's 15, 108, 9.

02:51 Change in mass, we will have to take 5 .00 mv, divided by c squared in terms of m mv per u, which is 931 .4944 m mv per u.

03:06 This change in mass is actually 0, 537, 7.

03:22 Now we know that energy is being released.

03:31 So the final products must have a smaller mass...

Please give Ace some feedback