Question

A gold nucleus has a radius of 7.3 \(\times 10^{-15}\) m and a charge of \(+79e\). Through what voltage must an alpha particle, with charge \(+2e\), be accelerated so that it has just enough energy to reach a distance of 2.0 \(\times 10^{-14}\) m from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

Short answer

The voltage required is approximately 8.34 MV.

Step-by-step solution

Understanding the Problem

We need to determine the voltage required to accelerate an alpha particle to a specific distance from a gold nucleus. The nucleus is treated as a point charge with a charge of \(+79e\). The initial and final distances from the center of the nucleus should be calculated in order to apply the concept of potential difference.

Calculating Initial and Final Distances

Since the alpha particle ends up at 2.0 \(\times 10^{-14}\) m from the surface of the gold nucleus, and the radius of the nucleus is 7.3 \(\times 10^{-15}\) m, the final distance from the center of the nucleus is \(r_f = 2.0 \times 10^{-14} + 7.3 \times 10^{-15} = 2.73 \times 10^{-14}\) m. The initial distance \(r_i\) is effectively infinite as the particle starts far away.

Applying Potential Energy Formula

The potential energy \(U\) of a charge \(q\) at a distance \(r\) from a point charge \(Q\) is given by \(U = \frac{kQq}{r}\). Substituting \(Q = +79e\) and \(q = +2e\), the potential energies at \(r_i\) and \(r_f\) are calculated.

Determining the Potential Difference

The potential difference (voltage) needed to achieve the desired separation can be found using \(V = \frac{kQ}{r}\). The change in potential energy \(\Delta U = U_f - U_i\) must equal the kinetic energy imparted to the alpha particle.

Calculating the Voltage

Using \(e = 1.6 \times 10^{-19}\) C and \(k = 8.99 \times 10^9\) N m²/C², calculate the voltage:\[V = \frac{k \cdot 79e \cdot 2e}{2.73 \times 10^{-14}}\]This gives \[V = \frac{(8.99 \times 10^9) \, (79) \, (2) \, (1.6 \times 10^{-19})^2}{2.73 \times 10^{-14}} \approx 8.34 \, \text{M}V\]

Conclusion

The required voltage is approximately 8.34 MV. This means the alpha particle must be accelerated through a potential difference of 8.34 million volts to reach the specified distance from the gold nucleus.

Key concepts

Electric Potential Energy

Electric potential energy is a form of energy associated with the position of a charged particle in an electric field.
This concept is integral to understanding how alpha particles interact with gold nuclei. When a charge is moved within an electric field, it gains or loses electric potential energy.
The potential energy, denoted as \( U \), between two point charges is calculated using the formula:

• \( U = \frac{kQq}{r} \), where:
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/ ext{C}^2) \)
• \( Q \) and \( q \) represent the charges
• \( r \) is the distance between the charges.

This energy concept answers why a voltage is needed to move an alpha particle to a specific point relative to another charged entity such as a gold nucleus.
It's the difference in electrical potential that contributes to the energy required for movement through an electric field.

Point Charge

A point charge is an idealized model where a charged object is assumed to be concentrated at a single point in space.
This simplification makes it easier to analyze the interactions between charges. In your exercise, the gold nucleus is treated as a point charge despite its actual physical size. Treating larger objects as point charges allows us to use simple equations, like Coulomb’s Law, without accounting for specific charge distribution. It also makes calculations manageable, especially when dealing with very small distances or very high charges.This implies that the gold nucleus, with a charge of \(+79e\), acts like a single concentrated charge that influences other charges, such as the \(+2e\) of an alpha particle, according to the rules of electrostatic interaction.
This assumption holds true mostly under cases where the position of interest is significantly farther from the nucleus compared to its size.

Alpha Particle

An alpha particle is a type of ionizing particle that consists of two protons and two neutrons bound together.
It is positively charged due to its pair of protons, carrying a charge of \(+2e\). This is a key component in nuclear reactions and radioactive decay processes.In this exercise, the alpha particle's positive charge makes it repelled by the similarly charged gold nucleus.
Hence, to overcome this repulsive force and bring the alpha particle close to the gold nucleus, the particle must be accelerated. Alpha particles are considerably more massive than electrons, meaning they require larger energies or voltages to achieve high velocities.
For a nuclear physics computation, accounting for their interaction with nuclear fields like that of the gold nucleus is essential.

Kinetic Energy

Kinetic energy is the energy that a body possesses due to its motion. In classical mechanics, this is given by:

• \( KE = \frac{1}{2}mv^2 \)
• Where \( m \) is mass
• \( v \) is velocity.

For charged particles, kinetic energy can also arise from changes in electric potential energy. In this exercise, the alpha particle's kinetic energy comes from the work done by the electric field of the gold nucleus. When an alpha particle accelerates through an electric field created by a potential difference (voltage), its potential energy converts into kinetic energy.
This conversion allows the particle to reach closer distances to the gold nucleus, overcoming the natural repulsive forces between positively charged entities.
Ultimately, the amount of kinetic energy gained by the particle is dictated by the potential difference applied, thus influencing how close it can approach another charged body like a gold nucleus.