91Ó°ÊÓ

A potential difference, also known as voltage, is a measure of the work needed to move a charge from one point to another in an electric field. It is denoted by the symbol \(V\) and typically measured in volts (V). In the context of this exercise, we must calculate the potential difference required to impart a specific kinetic energy to a helium nucleus.

The relationship between potential difference and kinetic energy in this scenario is given by the formula:

• \( E_k = QV \)

where \(E_k\) is the kinetic energy, \(Q\) is the charge of the helium nucleus, and \(V\) is the potential difference. Thus, the potential difference is crucial in this energy transfer because it determines how much kinetic energy can be gained by the helium nucleus as it traverses through the electric field.

By calculating the potential difference, we are effectively assessing how much energy per unit charge is available to be transferred to the charged particle, altering its kinetic state.

Kinetic energy is the energy of motion. It can be understood as the energy required to accelerate an object from the rest to its current velocity. In our exercise, the helium nucleus is accelerated to achieve a kinetic energy of 85.0 keV (kiloelectronvolts).

In physical contexts:

• \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ joules}\)

Converting kinetic energy from keV to joules makes it easier to work in standard SI units in calculations. In this problem, converting 85.0 keV to joules results in \(1.3617 \times 10^{-14} \text{ J}\).

Understanding kinetic energy allows us to see how potential differences contribute to altering the velocity and movement energy of particles such as the helium nucleus within electric fields.

A helium nucleus, also known as an alpha particle, consists of two protons and two neutrons. This structure gives it a specific set of properties, particularly in electrostatic interactions.

The charge of a helium nucleus is significant in numerical terms:

• \(Q = 2e = 3.204 \times 10^{-19} \text{ C}\)

where \(e\) is the elementary charge, approximately \(1.602 \times 10^{-19} \text{ C}\). The significance of the helium nucleus's charge lies in how it influences the dynamics as it moves within an electric field. The total charge impacts the force experienced by the particle due to electric potentials and thus affects the kinetic energy gained or lost during motion.

The conservation of energy is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of this exercise involving a helium nucleus, this principle helps us understand why and how kinetic energy changes as it moves across a potential difference.

When dealing with charged particles:

• The sum of kinetic energy and potential energy remains constant.

This principle ensures that any decrease in the potential energy of the nucleus is directly associated with an increase in its kinetic energy, as described by the equation \(E_k = QV\).

Therefore, conservation of energy is invaluable for predicting the outcomes of energy transformations within electrostatic systems and ensures that the calculated potential difference correctly reflects the energy dynamics described in the original exercise.