An alpha particle ( \(^{4} \mathrm{He}\) ) strikes a stationary gold nucleus \(\left(^{197} \mathrm{Au}\right)\) head-on. What fraction of the alpha's kinetic energy is transferred to the gold? Assume a totally elastic collision.

One of the fundamental principles in physics is the conservation of momentum. It states that within a closed system, with no external forces acting upon it, the total momentum remains constant over time. To understand this concept, picture two skaters on an ice rink – when they push off each other, they move in opposite directions, but the overall momentum of both before and after the push remains unchanged.

In the context of our alpha particle and gold nucleus collision, the alpha particle's initial forward momentum is equal to the combined momentum of both particles after the collision because the system is isolated and no external forces are present. This is mathematically expressed as \(m_1 \cdot v_{1i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f}\), where \(m_1\) and \(m_2\) represent the masses and \(v_{1i}\), \(v_{1f}\), and \(v_{2f}\) represent the initial and final velocities respectively.

In any elastic collision, not only is momentum conserved, but the total kinetic energy of the system is also conserved. Kinetic energy, which is the energy an object has due to its motion, is given by the formula \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity.

For our example, before collision, only the alpha particle has kinetic energy as the gold nucleus is at rest. After the collision, both particles move, and their total kinetic energy must equal the alpha particle's initial kinetic energy. This is expressed as \(\frac{1}{2}m_1 \cdot v_{1i}^2 = \frac{1}{2}m_1 \cdot v_{1f}^2 + \frac{1}{2}m_2 \cdot v_{2f}^2\). The challenge here is to use this principle, along with the conservation of momentum, to solve for the final velocities and then calculate the fraction of kinetic energy transferred.

In the vast world of atomic and subatomic particles, it's essential to have a standard unit of mass for comparison, and this is where the atomic mass unit (u) comes into play. It is defined as one twelfth the mass of a carbon-12 atom, approximately \(1.66053906660 \times 10^{-27}\) kilograms.

It's a tiny number, incredibly convenient for calculations in the atomic scale. In our problem, the masses of the alpha particle as \(4 u\) and gold nucleus as \(197 u\) are expressed in atomic mass units, highlighting the difference not only in magnitude but also emphasizing the practicality of using a standardized unit when dealing with such small scales.

What exactly is an alpha particle? An alpha particle, symbolized as \(^{4}\mathrm{He}\), is essentially a helium nucleus that consists of two protons and two neutrons. It's a heavy and positively charged particle emitted in the radioactive decay of some elements.

Because of its mass and charge, an alpha particle interacts with other atoms and particles in distinctive ways. When it collides with a gold nucleus in our problem, the interaction is an elastic collision due to the nature of forces during the collision that conserve both momentum and kinetic energy. Alpha particles are widely used in experiments to probe the structure of atoms, as seen in the famous gold foil experiment by Rutherford, leading to the discovery of the atomic nucleus.