Question

The potential difference between two parallel conducting plates in vacuum is 350 V. An alpha particle with mass of 6.50 x10-27 kg and charge of 3.20 x10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 27.0 cm.

Answer 1

Concepts and reason

The concept used to solve this problem is kinetic energy of the particle, work done on a charged particle in an electric field, and work-energy theorem.

Initially, calculate the work done on the alpha particle using work done on a charged particle in an electric field, then find the change in kinetic energy of the alpha particle using the expression of kinetic energy and finally according to the work-energy theorem equate this change in kinetic energy to the work done to find final kinetic energy.

Fundamentals

The work done on a charged particle in moving a particle in electric field is,

$W = \Delta V \cdot q$

Here, W is the work done, $\Delta V$ is the potential difference, and q is the charge of the particle.

According to Work-Energy theorem, work done is equal to the change in the kinetic energy.

$W = \Delta K$

Here, $\Delta K$ is the change in the kinetic energy.

The kinetic energy of the particle is,

$K = \frac{1}{2}m{v^2}$

Here, K is the kinetic energy, m is the mass of the particle, and v is the velocity of the particle.

Calculate the work done on the alpha particle as follows:

The work done on a charged particle in moving a particle in electric field is,

$W = \Delta V \cdot q$

Substitute 350 V for $\Delta V$ and $3.20 \times {10^{ - 19}}{\rm{ C}}$ for q.

$\begin{array}{c}\\W = \left( {350{\rm{ V}}} \right) \cdot \left( {3.20 \times {{10}^{ - 19}}{\rm{ C}}} \right)\\\\ = 1.12 \times {10^{ - 16}}{\rm{ J}}\\\end{array}$

Calculate the change in kinetic energy of the alpha particle as follows:

The initial kinetic energy (${K_i}$) of the particle is,

${K_i} = \frac{1}{2}m{v_i}^2$

Here, m is the mass of the particle and ${v_i}$ is the initial velocity.

Substitute 0 m/s for ${v_i}$.

$\begin{array}{c}\\{K_i} = \frac{1}{2}m{\left( {0{\rm{ m/s}}} \right)^2}\\\\ = 0{\rm{ J}}\\\end{array}$

The change in kinetic energy ($\Delta K$) of the particle is,

$\Delta K = {K_f} - {K_i}$

Here, ${K_f}$ is the final kinetic energy.

Substitute 0 J for ${K_i}$.

$\begin{array}{c}\\\Delta K = {K_f} - \left( {0{\rm{ J}}} \right)\\\\ = {K_f}\\\end{array}$

Calculate the final kinetic energy of the alpha particle as follows:

According to Work-Energy theorem, work done is equal to the change in the kinetic energy.

$W = \Delta K$

Here, $\Delta K$ is the change in the kinetic energy.

Substitute ${K_f}$ for $\Delta K$ and $1.12 \times {10^{ - 16}}{\rm{ J}}$ for W.

$\begin{array}{c}\\1.12 \times {10^{ - 16}}{\rm{ J}} = {K_f}\\\\{K_f} = 1.12 \times {10^{ - 16}}{\rm{ J}}\\\end{array}$

Ans:

The final kinetic energy of the particle is$1.12 \times {10^{ - 16}}{\rm{ J}}$.