Problem 63 - Physics

Question

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed \(v_{\mathrm{e}}\), while the proton acquires a speed \(v_{\mathrm{p}}\). (a) As each particle accelerates from rest, it gains kinetic energy. Does it gain or lose electric potential energy? (b) Does the electron gain more, less, or the same amount of kinetic energy as the proton does? (c) Is \(v_{\mathrm{e}}\) greater than, less than, or equal to \(v_{\mathrm{p}}\) ? Justify your answers. Find the ratio \(v_{\mathrm{e}} / v_{\mathrm{p}}\). Verify that your answer is consistent with your answers to the Concept Questions.

Step-by-Step Solution

Answer

Electrons lose potential and gain kinetic energy. Electrons gain the same kinetic energy but reach higher speeds than protons. The speed ratio is approximately 42.85.

Step 1: Understanding Electric Potential Energy and Kinetic Energy

When a charged particle is accelerated through an electric potential difference \(V\), the change in electric potential energy is converted to kinetic energy. An electron or proton gains kinetic energy while moving from high to low potential, thus losing electric potential energy.

Step 2: Energy Conservation Equation

The gain in kinetic energy for a particle is given by \(qV\) where \(q\) is the charge of the particle. Thus, the kinetic energy gained (KE) for both particles when accelerated through the same potential difference is: \[ KE = qV \] Both gain the same energy due to equal magnitude of their charge (ignoring the negative sign for electron's charge).

Step 3: Comparing Kinetic Energies

Since both particles have the same charge magnitude and travel through the same potential difference, they gain the same amount of kinetic energy. The energy gained is \(qV\) for both.

Step 4: Relating Mass and Speed

The expression for kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \] Equating the kinetic energies for the electron (\(e\)) and the proton (\(p\)) gives: \[ \frac{1}{2} m_{ ext{e}} v_{ ext{e}}^2 = \frac{1}{2} m_{ ext{p}} v_{ ext{p}}^2 \] Where \(m_{ ext{e}}\) and \(m_{ ext{p}}\) are the masses of the electron and proton, respectively.

Step 5: Solve for the Speed Ratio

From the equation\[ m_{ ext{e}} v_{ ext{e}}^2 = m_{ ext{p}} v_{ ext{p}}^2 \], extract the speed ratio: \[ \left( \frac{v_{ ext{e}}}{v_{ ext{p}}} \right)^2 = \frac{m_{ ext{p}}}{m_{ ext{e}}} \] Thus: \[ \frac{v_{ ext{e}}}{v_{ ext{p}}} = \sqrt{\frac{m_{ ext{p}}}{m_{ ext{e}}}} \] Given that \(m_{ ext{p}} \approx 1836 m_{ ext{e}}\), the ratio is: \[ \frac{v_{ ext{e}}}{v_{ ext{p}}} = \sqrt{1836} \approx 42.85 \]

Step 6: Interpret the Result

The electron acquires a much higher speed than the proton due to its significantly smaller mass. This result confirms that \(v_{ ext{e}} > v_{ ext{p}}\) because the mass of the proton is much greater than that of the electron.

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. When particles like electrons and protons start moving, they naturally gain kinetic energy. In the context of our example, an electron and a proton are initially at rest, meaning they have no kinetic energy. When they are accelerated by an electric potential difference, they start moving and gain kinetic energy. The formula for kinetic energy is:

• \[ KE = \frac{1}{2} mv^2 \]

Here, \(m\) represents the mass of the particle, and \(v\) is the velocity it gains while moving. Since both the electron and proton are exposed to the same potential difference, the kinetic energy they gain, calculated by \(qV\), remains the same though their speeds differ due to mass differences.

Electric Potential Energy

Electric potential energy is the energy stored due to an object's position in an electric field. This form of energy transforms when charged particles like electrons or protons move through an electric potential difference, \(V\). As they move from areas of high potential to low potential, they lose electric potential energy and gain kinetic energy. For particles with the same charge magnitude like protons and electrons, the change in electric potential energy can be expressed by:

• \[ \Delta PE = qV \]

Here, \(q\) is the charge of the electron or proton, and \(V\) is the potential difference they are subjected to. Unlike kinetic energy, electric potential energy decreases as the charged particles start to move. This decrease balances out as a gain in kinetic energy.

Mass-Energy Equivalence

Mass-energy equivalence, famously captured by Einstein's formula \( E = mc^2 \), refers to the way mass can be converted into energy and vice versa. Although this concept is rooted deeply in relativity, it also helps us understand the particle behavior in electric fields. When an electron and a proton gain kinetic energy, the differences in their masses dramatically affect their velocities. Since the proton's mass is approximately 1836 times more than that of an electron, when both gain the same kinetic energy (\(qV\)), the electron ends up moving much faster than the proton. This is because the speed gained by these particles is inversely proportional to the square root of their masses, as expressed in the equation:

• \[ \frac{v_e}{v_p} = \sqrt{\frac{m_p}{m_e}} \]

This abstraction neatly links mass productions with their resultant speeds, further emphasizing the principle of mass-energy equivalence in dynamic conditions.