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Proton acceleration occurs when a proton (\(q = e, \ m_p = 1.67 \times 10^{-27}\text{kg}\)) is subjected to an electric field, gaining speed and kinetic energy. When we apply a potential difference to a proton, the force exerted by the electric field accelerates the proton from rest. This force changes the proton's speed by converting potential energy into kinetic energy. The equation \(F = qE\) cannot be ignored, but for potential difference calculations, we focus on energy transformations.

• Acceleration means an increase in velocity over time.
• An electric field causes the proton to accelerate.
• Force is a product of charge and electric field \(F = qE\).

Therefore, understanding the proton's response in terms of energy is more manageable when the potential difference is the main factor.

The concept of potential difference, often referred to as voltage, is crucial in understanding how protons are accelerated. A potential difference, measured in volts (V), represents the work done per unit charge to move a charge between two points.
When a proton is accelerated by a potential difference, it gains energy. For a potential difference \(V\) of \(1.0 \text{ MV} = 1.0 \times 10^6 \text{ V}\), the proton gains energy equal to its charge \(q\) multiplied by the potential difference \(V\): \(E = qV\). For a proton, \(q = 1.6 \times 10^{-19}\text{C}\), so the energy gained is given by:
\[E = 1.6 \times 10^{-19} \text{ C} \times 1.0 \times 10^6 \text{ V} = 1.6 \times 10^{-13} \text{ J}\]

• Potential difference is measured in volts (V).
• 1 MV equals \(1.0 \times 10^6\) volts.
• The proton gains energy when crossing this potential difference.

Understanding this concept allows us to calculate how much energy a proton will gain.

Kinetic energy is the energy of motion. When a proton is accelerated through a potential difference, its electric potential energy converts into kinetic energy. Kinetic energy (KE) is calculated using the formula,\(KE = \frac{1}{2} mv^2\), where \(m\) is the mass of the proton. Substituting in the provided data,

• Mass of proton, \(m_p = 1.67 \times 10^{-27}\text{ kg}\)
• Kinetic energy gained \(E = 1.6 \times 10^{-13}\text{ J}\)

The gained kinetic energy equals the energy calculated from potential difference:\(E = \frac{1}{2} m v^2 = 1.6 \times 10^{-13} \text{ J}\). Solving for \(v^2\) and then for \(v\) provides insight into the speed the proton acquires. This connection between energy forms plays a vital role in understanding motion in electric fields.

To determine the velocity of a proton after acceleration, we use the calculated kinetic energy. Start with the formula\[\frac{1}{2} m v^2 = 1.6 \times 10^{-13} \text{ J}\].Rearranging the formula for \(v^2\):\[v^2 = \frac{2 \times 1.6 \times 10^{-13} \text{ J}}{1.67 \times 10^{-27} \text{ kg}}\],we find that\(v^2 = \frac{3.2 \times 10^{-13}}{1.67 \times 10^{-27}} \approx 1.917 \times 10^{14} \). To find the velocity \(v\), take the square root:\[v = \sqrt{1.917 \times 10^{14}} \approx 1.38 \times 10^7 \text{ m/s}\].This result shows the velocity of a proton after being accelerated by a potential difference of 1.0 MV. The relationship between the potential difference, energy, and velocity clearly illustrates the conversion of potential to kinetic energy in an easily measurable velocity.