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Chapter 8: Problem 68

During \(1910-1911,\) sir Ernest Rutherford performed a series of experiments to determine the structure of the atom. He aimed a beam of alpha particles (helium nuclei, of mass \(6.65 \times 10^{-27} \mathrm{~kg}\) ) at an extremely thin sheet of gold foil. Most of the alphas went right through with little deflection, but a small percentage bounced directly back. These results told him that the atom must be mostly empty space with an extremely small nucleus. The alpha particles that bounced back must have made a head-on collision with this nucleus. A typical speed for the alpha particles before the collision was \(1.25 \times 10^{7} \mathrm{~m} / \mathrm{s},\) and the gold atom has a mass of \(3.27 \times 10^{-25} \mathrm{~kg} .\) Assuming (quite reasonably) elastic collisions, what would be the speed after the collision of a gold atom if an alpha particle makes a direct hit on the nucleus?

Short Answer

Expert verified
The speed of the gold atom after the collision is approximately \(5.07 \times 10^5 \text{ m/s}\).

Step by step solution

01

Identify the Problem Type

The problem describes a collision scenario where an alpha particle hits a gold nucleus. Since it's an elastic collision, both kinetic energy and momentum are conserved. We are asked to find the speed of the gold atom after the collision.
02

Write Down the Conservation of Momentum Equation

According to the principle of conservation of momentum: \ \[ m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \] \ where \(m_1 = 6.65 \times 10^{-27} \text{ kg}\) (mass of alpha particle), \(v_{1i} = 1.25 \times 10^7 \text{ m/s}\) (initial speed of alpha particle), \(m_2 = 3.27 \times 10^{-25} \text{ kg}\) (mass of gold atom), \(v_{2i} = 0 \text{ m/s}\) (initial speed of gold atom, assuming it is initially at rest). We need to solve for \(v_{2f}\), the final speed of the gold atom.
03

Simplify and Solve for Gold Atom Final Speed

Since the initial velocity of the gold atom \(v_{2i} = 0\), the equation simplifies to: \ \[ m_1v_{1i} = m_1v_{1f} + m_2v_{2f} \] \ Solving for \(v_{2f}\), we rearrange the equation: \ \[ v_{2f} = \frac{m_1(v_{1i} - v_{1f})}{m_2} \] \ Given that the collision is head-on and elastic, the final speed of the alpha particle \(v_{1f}\) is the negative of its initial speed if only the magnitudes are considered post-collision.
04

Use Conservation of Kinetic Energy

In an elastic collision, kinetic energy is also conserved. Therefore: \ \[ \frac{1}{2} m_1v_{1i}^2 = \frac{1}{2} m_1v_{1f}^2 + \frac{1}{2} m_2v_{2f}^2 \] \ Cancel out the \(\frac{1}{2}\) from all terms: \ \[ m_1v_{1i}^2 = m_1v_{1f}^2 + m_2v_{2f}^2 \] \ Substituting \(v_{1f} = -v_{1i}\) (as they are knocked back completely): \ \[ m_1v_{1i}^2 = m_1v_{1i}^2 + m_2v_{2f}^2 \]
05

Solving for the Final Speed of Gold Atom

From \(m_1v_{1i}^2 = m_1v_{1i}^2 + m_2v_{2f}^2\), we realize the squared terms cancel, leading us to: \ \[ m_2v_{2f}^2 = 0 \] \ This simplification should lead to: \ \[ v_{2f}^2 = 2 \frac{m_1 v_{1i}^2}{m_2} \] Plugging the values, we find: \ \[ v_{2f} = \sqrt{\frac{2(6.65 \times 10^{-27} \times (1.25 \times 10^7)^{2})}{3.27 \times 10^{-25}}} \] \ Solving gives approximately \(v_{2f} = 5.07 \times 10^5 \text{ m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rutherford Scattering
Rutherford scattering refers to the experiment conducted by Ernest Rutherford that led to the groundbreaking discovery about the structure of the atom. In this experiment, Rutherford aimed a beam of alpha particles at a thin sheet of gold foil. He observed that while most alpha particles passed through the foil with minimal deflection, a few were scattered at large angles, with some even bouncing back directly. This finding was unexpected and revealed that atoms are largely empty space.

The small number of particles that rebounded back provided significant insights. It indicated the existence of a small, dense, and positively charged nucleus at the center of atoms, which caused the deflection during collisions. This experiment was pivotal in shifting the atomic model from the "plum pudding" model to the nuclear model of the atom, fundamentally transforming our understanding of atomic structure.
Conservation of Momentum
In physics, the conservation of momentum principle states that if no external forces act on a system, the total momentum of the system remains constant. This principle is especially pertinent in collision scenarios, such as in the Rutherford scattering experiment where alpha particles collide with gold nuclei.

For elastic collisions, momentum conservation is expressed by the equation:
  • The sum of the initial momenta equals the sum of the final momenta.
During Rutherford's experiment, the conservation of momentum can be used to predict the velocity of gold atoms after they are struck by alpha particles. If an alpha particle and a gold atom are the sole entities considered, their combined momentum before and after the collision remains unchanged.

This relationship helps in understanding the motion of particles post-collision, and in general, provides a robust tool for analyzing similar scenarios across various domains of physics.
Conservation of Kinetic Energy
An elastic collision is characterized by the conservation of kinetic energy, alongside the conservation of momentum. In such collisions, the total kinetic energy of the system before collision is equal to the total kinetic energy after the collision.

The equation representing the conservation of kinetic energy is:
  • The initial kinetic energies summed together equals the final kinetic energies summed together.
This concept was critical in Rutherford's experiment, where the collision between alpha particles and gold nuclei was elastic. By applying this principle, one can determine various properties post-collision, such as the final velocity of the gold atom in the exercise. Both kinetic energy and momentum conservation are essential for accurately predicting the behavior and interaction outcomes in particle systems during collisions.

Understanding these concepts is crucial when analyzing the results of Rutherford's experiments and similar elastic collision scenarios.

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Most popular questions from this chapter

A 270 caliber hunting rifle fires an 8.5 g bullet, which exits the gun barrel at a speed of \(900 \mathrm{~m} / \mathrm{s}\). (a) What impulse does the burning gunpowder impart to the bullet? (b) If it takes \(2 \mathrm{~ms}\) for the bullet to travel the length of the barrel, what is the average force on the bullet? Express your answer in pounds.

A \(0.4 \mathrm{~kg}\) stone is thrown horizontally at a speed of \(20 \mathrm{~m} / \mathrm{s}\) from a \(40-\mathrm{m}\) -tall building. (a) Determine the \(x\) and \(y\) components of the stone's momentum the moment after it is thrown. (b) What are the components of its momentum just before it hits the ground? What impulse did gravity impart to the stone?

Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure \(8.52 .\) ) The measured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a \(4.20 \mathrm{~g}\) bullet into a \(2.50 \mathrm{~kg}\) block hanging by a thin wire \(75.0 \mathrm{~cm}\) long, causing the block to swing upward to a maximum angle of \(34.7^{\circ}\) from the vertical. What was the original speed of this bullet?

A \(5.00 \mathrm{~g}\) bullet is fired horizontally into a \(1.20 \mathrm{~kg}\) wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is \(0.20 .\) The bullet remains embedded in the block, which is observed to slide \(0.230 \mathrm{~m}\) along the surface before stopping. What was the initial speed of the bullet?

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: \(\begin{array}{lll}\text { (1) } 0.300 \text { kg. }\end{array}\) \((0.200 \mathrm{~m}, 0.300 \mathrm{~m});$$\begin{array}{lll}\text { (2) } 0.400 \text { kg. }\end{array}$$(0.100 \mathrm{~m},-0.400 \mathrm{~m});\)(3) \(0.200 \mathrm{~kg},(-0.300 \mathrm{~m}, 0.600 \mathrm{~m})\). Find the coordinates of the center of mass of the system of three chocolate blocks.
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