Structure of an Atom In 1911 , Ernest Rutherford discovered...

Structure of an Atom In 1911 , Ernest Rutherford discovered the basic structure of the atom by "shooting" positively charged alpha particles with a speed of $10^{7}$ meters per second at a piece of gold foil $6 \times 10^{-7}$ meter thick. Only a small percentage of the alpha particles struck a gold nucleus head-on and were deflected directly back toward their source. The rest of the particles often followed a hyperbolic trajectory because they were repelled by positively charged gold nuclei. Thus, Rutherford proposed that the atom was composed of mostly empty space and a small, dense nucleus. The figure shows an alpha particle $A$ initially approaching a gold nucleus $N$ and being deflected at an angle $\theta=90^{\circ}$ $N$ is located at a focus of the hyperbola, and the trajectory of $A$ passes through a vertex of the hyperbola. (Source: Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.) (a) Determine the equation of the trajectory of the alpha particle if $d=5 \times 10^{-14}$ meter. (b) Approximate the minimum distance between the centers of the alpha particle and the gold nucleus.

Structure of an Atom In 1911 , Ernest Rutherford discovered the basic structure of the atom by "shooting" positively charged alpha particles with a speed of $10^{7}$ meters per second at a piece of gold foil $6 \times 10^{-7}$ meter thick. Only a small percentage of the alpha particles struck a gold nucleus head-on and were deflected directly back toward their source. The rest of the particles often followed a hyperbolic trajectory because they were repelled by positively charged gold nuclei. Thus, Rutherford proposed that the atom was composed of mostly empty space and a small, dense nucleus. The figure shows an alpha particle $A$ initially approaching a gold nucleus $N$ and being deflected at an angle $\theta=90^{\circ}$ $N$ is located at a focus of the hyperbola, and the trajectory of $A$ passes through a vertex of the hyperbola. (Source:
Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.)
(a) Determine the equation of the trajectory of the alpha particle if $d=5 \times 10^{-14}$ meter.
(b) Approximate the minimum distance between the centers of the alpha particle and the gold nucleus.

Key Concepts

Rutherford Scattering

This concept refers to the experiment in which charged particles, such as alpha particles, are directed at thin metal foils to probe the internal structure of atoms. The observation that most particles passed through with little deflection while a few were significantly scattered led to the discovery that atoms consist of mostly empty space with a small, dense nucleus. This experiment was pivotal in shifting the understanding of atomic structure away from the ‘plum pudding’ model.

Atomic Nucleus

The atomic nucleus is the small, dense region at the center of an atom that contains most of its mass and positive charge. Its discovery through scattering experiments revealed that electrons occupy a vast space around the nucleus, fundamentally influencing the modern understanding of chemical behavior and atomic interactions.

Coulomb Interaction

Coulomb interaction refers to the force between two charged particles, which in the context of scattering experiments is a repulsive force between the positively charged alpha particles and the nucleus. This force, governed by Coulomb's law, dictates the acceleration and trajectory of the particles during their interaction, and is essential for calculating deflection paths.

Hyperbolic Trajectory

When a charged particle such as an alpha particle is influenced by the Coulomb force in a repulsive interaction, its path can be described by a hyperbola. Hyperbolic trajectories arise in systems where particles approach a scattering center and then escape to infinity, and their mathematical description involves conic sections that are derived from the conservation principles of energy and momentum.

Conservation Laws

The conservation of energy and angular momentum are fundamental principles used to analyze and derive the motion of particles under the influence of forces. In the context of scattering problems, these laws allow for the determination of trajectory equations and minimum approach distances by relating the kinetic energy and potential energy arising from Coulomb interactions.

Transcript

00:01 For this problem, we're going to start by finding the equation of the hyperbola that matches our picture given.

00:09 Because it's centered at the origin, my numerators are just going to be x squared and y squared.

00:15 And i know the x squared comes first because i can see that the major axis parallels the x -axis.

00:20 My denominator will be a squared and b squared.

00:25 Now the problem is i don't know either a or b here, but i do know something about how they relate to each other.

00:30 I know that my two asymptopes are y equals x and y equals negative x.

00:38 That tells me that my slope, b over a, is going to equal 1 or b equals a.

00:45 So i'm just going to put that in parentheses here.

00:47 We will refer back to that in just a moment.

00:50 What the number that i am given is d.

00:53 I don't have a or b, but i do have d.

00:55 So let's take a look at that small triangle that's shown on our graph because that tells us where d is in really, to everything else.

01:04 The one vertex is n.

01:06 The vertex i'm marking in red is the origin or center of the hyperbola at 0 -0.

01:11 And i know that this side of the triangle is d.

01:15 Well, because i know that the asymptopes are the lines y equals x and y equals negative x, this is a 45 -4590 triangle, which means that the sides of my triangles are d, d, and square root of 2 times d.

01:30 That distance from n to the origin, square root of 2 times d, that equals c, the distance from the focus to the center.

01:39 And i know for any hyperbola, c squared equals a squared plus b squared.

01:47 A and b are equal.

01:50 So i can say that c squared equals to a squared.

01:54 And i know that c equals square root of 2 times d.

01:57 So i can change this to say 2d squared equals 2a squared.

02:02 Or d equals a.

02:05 D also equals b, since b equals a.

02:08 So now i can finish writing the equation of my hyperbola.

02:12 X squared over a squared...