Question

A \(6.31-\mathrm{MeV}\) alpha particle \(\left(\right.\) mass \(\left.=3.7274 \mathrm{GeV} / \mathrm{c}^{2}\right)\) inside a heavy nucleus encounters a barrier whose average height is \(15.7 \mathrm{MeV}\) and whose width is \(13.7 \mathrm{fm}\left(1 \mathrm{fm}=1 \cdot 10^{-15} \mathrm{~m}\right)\). What is the value of the decay constant, \(\gamma\left(\right.\right.\) in \(\left.\left.\mathrm{fm}^{-1}\right)\) ? (Hint: A potentially useful value is \(\left.\hbar c=197.327 \mathrm{MeV} \mathrm{fm} .\right)\)

Short answer

Question: Calculate the decay constant for a 6.31 MeV alpha particle encountering a barrier with an average height of 15.7 MeV and a width of 13.7 fm. Answer: The decay constant for this problem is approximately \(0.1436 \ \mathrm{fm}^{-1}\).

Step-by-step solution

Write down the given information

The information given in the problem is: 1. Alpha particle energy: \(E = 6.31 \ \mathrm{MeV}\) 2. Barrier height: \(V_{0} = 15.7 \ \mathrm{MeV}\) 3. Barrier width: \(d = 13.7 \ \mathrm{fm}\) 4. Mass of alpha particle: \(m = 3.7274 \ \mathrm{GeV}/c^2\) 5. \(\hbar c = 197.327 \ \mathrm{MeV} \ \mathrm{fm}\)

Calculate the decay constant, \(\gamma\)

We first need to calculate the difference between the barrier height and the energy of the particle, which represents the potential energy. \(V = V_{0} - E = 15.7\ \mathrm{MeV} - 6.31\ \mathrm{MeV} = 9.39\ \mathrm{MeV}\) To find the decay constant, \(\gamma\), we use the formula: \(\gamma = \sqrt{\dfrac{2mV}{\hbar^2}}\) However, we need to convert mass of the alpha particle to \(\mathrm{MeV}/c^2\) to make sure we are using the appropriate units.

Convert mass to \(\mathrm{MeV}/c^2\)

Since \(1\ \mathrm{GeV} = 1000\ \mathrm{MeV}\), we can convert the mass to \(\mathrm{MeV}/c^2\) as follows: \(m_\text{converted} = 3.7274 \ \mathrm{GeV}/c^2 \cdot \dfrac{1000\ \mathrm{MeV}}{1 \ \mathrm{GeV}} = 3727.4\ \mathrm{MeV}/c^2\) We can now plug this converted mass value into the formula to find \(\gamma\).

Find the decay constant \(\gamma\)

Using the provided value for \(\hbar c = 197.327\ \mathrm{MeV} \ \mathrm{fm}\), we have: \(\gamma = \sqrt{\dfrac{2m_\text{converted}V}{\hbar^2 c^2}} = \sqrt{\dfrac{2(3727.4)(9.39)}{(197.327)^2}}\) Solving for \(\gamma\): \(\gamma \approx 0.1436 \ \mathrm{fm}^{-1}\) So the decay constant for this problem is approximately \(0.1436 \ \mathrm{fm}^{-1}\).

Key concepts

Decay Constant

The decay constant, often represented by the Greek letter \( \gamma \), is a fundamental parameter in nuclear physics that describes the rate at which an unstable nucleus will undergo radioactive decay. It is inversely related to the half-life of the radioactive substance, with a larger decay constant indicating a faster rate of decay.

Understanding the decay constant is crucial when we analyze quantum tunneling, the phenomenon where a particle penetrates a potential barrier, such as an alpha particle escaping a nucleus. The decay constant in the context of quantum mechanics is related to the probability of a particle tunneling through a barrier and can be calculated using the alpha particle's energy \(E\), the height of the barrier \(V_0\), the particle's mass \(m\), and Planck's constant \(\hbar\).

The formula involves the square root of the product of the particle's mass, the potential energy difference \( (V_0 - E) \) and twice the reduced Planck's constant \(\hbar^2\). The calculation yields a value for the decay constant \(\gamma\) in inverse femtometers \(\mathrm{fm}^{-1}\) which gives us an indication of how likely or quickly the alpha particle is to tunnel through the nuclear barrier.

Alpha Particle

An alpha particle is a type of ionizing radiation ejected from the nuclei of certain radioactive atoms during the process of alpha decay. Alpha particles consist of two protons and two neutrons, which is the same as a helium-4 nucleus.

These particles are a benchmark example when studying nuclear physics, as they frequently illustrate concepts such as nuclear decay, the strong force, and quantum tunneling. Despite their relatively large mass and charge, which typically render them less penetrating than other forms of radiation such as beta particles, alpha particles can sometimes escape the potential well of a heavy nucleus due to quantum tunneling.

In the given exercise, the escape of an alpha particle with an energy of \(6.31\ \mathrm{MeV}\) from a heavy nucleus through a potential barrier was analyzed. This process, which is counterintuitive from a classical physics standpoint, is made possible because subatomic particles exhibit wave-particle duality and can, under certain conditions, leak out of a nuclear well they ostensibly don't have the energy to surmount.

Nuclear Physics

Nuclear physics is a field within physics that explores the constituents and interactions of atomic nuclei. Central to this field is the study of nuclear forces, reactions, structures, and the fascinating phenomena that stem from the properties of the nuclear world, such as fusion, fission, and radioactive decay.

Quantum tunneling, a prominent concept in nuclear physics, allows particles to bypass a classically insurmountable energy barrier through quantum effects, as found in the decay of atomic nuclei. This aspect of quantum mechanics is pivotal in understanding our exercise, where an alpha particle tunnels through a barrier higher than its own kinetic energy. The rigorous analysis of these interactions not only requires knowledge of the quantum mechanics involved but also an understanding of how various constants, like the decay constant and Planck’s constant, come into play.

Nuclear physicists employ these principles to explain and predict a wide range of phenomena, from the behavior of stars to the generation of nuclear power, thus underscoring the profound impact these microscopic factors have on our macroscopic world.