The first nuclear reaction ever observed (in 1919 by Ernest...

The first nuclear reaction ever observed (in 1919 by Ernest Rutherford) was $\alpha+{ }_{7}^{14} \mathrm{~N} \rightarrow \mathrm{p}+\mathrm{X} .$ (a) Show that the reaction product "X" must be ${ }_{8}^{17} \mathrm{O}$. (b) For this reaction to take place, the alpha particle must come in contact with the nitrogen nucleus. Calculate the distance $d$ between their centers when they just make contact. (c) If the alpha particle and the nitrogen nucleus are initially far apart, what is the minimum value of their kinetic energy necessary to bring the two into contact? Express your answer in terms of the elementary charge $e$, the contact distance $d$, and whatever else you need. (d) Is the total kinetic energy of the reaction products more or less than the initial kinetic energy in part (c)? Why? Calculate this kinetic energy difference.

The first nuclear reaction ever observed (in 1919 by Ernest Rutherford) was $\alpha+{ }_{7}^{14} \mathrm{~N} \rightarrow \mathrm{p}+\mathrm{X} .$ (a) Show that the reaction product "X" must be ${ }_{8}^{17} \mathrm{O}$. (b) For this reaction to take place, the alpha particle must come in contact with the nitrogen nucleus. Calculate the distance $d$ between their centers when they just make contact.
(c) If the alpha particle and the nitrogen nucleus are initially far apart, what is the minimum value of their kinetic energy necessary to bring the two into contact? Express your answer in terms of the elementary charge $e$, the contact distance $d$, and whatever else you need.
(d) Is the total kinetic energy of the reaction products more or less than the initial kinetic energy in part (c)? Why? Calculate this kinetic energy difference.

Key Concepts

Coulomb Barrier in Nuclear Reactions

Due to the positive charges in nuclei, there exists an electrostatic repulsion known as the Coulomb barrier. This barrier represents the potential energy that two approaching charged nuclei must overcome in order to come into contact and participate in a nuclear reaction. The minimum kinetic energy required for the reaction to occur is determined by the energy needed to overcome this electrostatic potential at the contact distance.

Q-Value and Kinetic Energy Considerations

The Q-value of a nuclear reaction quantifies the net energy released or absorbed during the reaction, based on the mass difference between reactants and products. In exothermic reactions, a positive Q-value signifies that some of the rest mass energy has been converted into kinetic energy of the reaction products. Energy conservation, therefore, implies that the total kinetic energy available after the reaction will differ from the initial kinetic energy required to overcome the Coulomb barrier, with the difference attributed to the energy released (or absorbed) as described by the Q-value.

Conservation Laws in Nuclear Reactions

In any nuclear reaction, the conservation of charge and nucleon number (mass number) holds true. This principle means that the total number of protons and neutrons must remain the same before and after the reaction, which is used to determine unknown reaction products by balancing both the charge (atomic numbers) and mass numbers in the reaction equation.

Nuclear Radii and the Contact Distance

Atomic nuclei can be modeled as spherical objects whose radii scale with the cube root of their mass number (A), often represented by the relation R = r0 * A^(1/3), where r0 is a constant derived from experimental data. When two nuclei approach one another, the minimum center-to-center distance, when their surfaces are just touching, is approximately the sum of their individual radii. This concept is essential for understanding nuclear collision geometry.

Transcript

00:01 For this question we are looking at the bombardment of alpha particles or nitrogen atom.

00:09 Right? and the equation is given above over here and we have an unknown daughter nucleic x and we'll find out what it is.

00:25 And to figure that out, we just need to look at the conservation of the proton number as well as the nucleon number.

00:32 So they must be the same for the left -hand side and the right -hand side.

00:35 They must be conserved.

00:37 On the left -hand side, we see that the total proton number is 9.

00:41 So on the right -hand side, we expect x to have 8 proton number, so 8 plus 1 is 9.

00:47 Same for the nuclear number.

00:49 On the left -hand side, total is 18.

00:52 So on the right -hand side, we should have 17 for the x, so that 17 plus 1 is 18.

01:00 Now we can look up the periodic table for our element with the right -hand side.

01:03 8 protons and that is oxygen.

01:07 So x must be oxygen 17.

01:14 Now for this reaction, we know that we are given that the alpha particle must be in contact with the nitrogen nuclei.

01:23 And in contact, right, this is what it means by in contact.

01:29 When this happens the distance between the two of the center of this nuclei distance d is equals to the radius of alpha plus the radius of nitrogen and this we can approximate using the formula r not times the nucleon number to the power of wanted we do this for both the alpha particle as well as the nitrogen so a alpha is four a of nitrogen, 14.

02:12 Substitute the values in.

02:15 If you should get, and also r0 is actually 1 .2, fermis.

02:24 If you substitute the values properly, we should get 4 .8 firmis.

02:33 Now we want to find what is the minimum ke for the alpha and the nitrogen to touch...

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