Question

(a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 158 V.
m/s

(b) Calculate the speed of an electron that is accelerated through the same potential difference.
m/s

Answer 1

Concepts and reason

Use the concepts of electric potential energy, kinetic energy, and conservation of energy to solve this problem.

First, find the speed of the protons that is accelerated from rest by using the concept of conservation of energy and electric potential energy.

Find the speed of the electron that is accelerated from rest by using the concept of conservation of energy and electric potential energy.

Fundamentals

The expression for the change in the electric potential energy is as follows:

$\Delta {U_E} = q\left( {\Delta V} \right)$ …… (1)

Here,$\Delta {U_E}$ is the change in the electric potential energy, q is the energy, and $\Delta V$is the potential difference.

The expression for the change in the kinetic energy is as follows:

$\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$ …… (2)

Here,$\Delta K$is the change in the kinetic energy, m is the mass of the sub atomic particle, ${v_f}$is the final velocity, and${v_i}$is the initial velocity.

Energy neither be created nor be destroyed but can be transformed from one form to other. This is called the conservation of energy.

From the conservation of energy, the change in the electric potential energy is equal to the change in the kinetic energy.

$\Delta {U_E} = \Delta K$ …… (3)

Substitute (1) and (2) in (3) as follows:

$\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = q\left( {\Delta V} \right)$ …… (4)

(a)

The expression for the conservation of energy from the equation (4) is,

$\frac{1}{2}{m_p}v_f^2 - \frac{1}{2}{m_p}v_i^2 = {q_p}\left( {\Delta V} \right)$

Here,${m_p}$is the mass of the proton and${q_p}$is the charge of the proton.

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

$\begin{array}{c}\\\frac{1}{2}{m_p}v_f^2 - \frac{1}{2}{m_p}\left( {0\;{\rm{m/s}}} \right) = {q_p}\left( {\Delta V} \right)\\\\\frac{1}{2}{m_p}v_f^2 = {q_p}\left( {\Delta V} \right)\\\end{array}$

Rearrange the above equation for${v_f}$.

The expression for the speed of the proton that is accelerated from rest is,

${v_f} = \sqrt {\frac{{2{q_p}\left( {\Delta V} \right)}}{{{m_p}}}}$

Substitute $\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)$for${q_p}$, 158 V for$\Delta V$, and $\left( {1.67 \times {{10}^{ - 27}}\;{\rm{kg}}} \right)$for${m_p}$.

$\begin{array}{c}\\{v_f} = \sqrt {\frac{{2\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {158\;{\rm{V}}} \right)}}{{\left( {1.67 \times {{10}^{ - 27}}\;{\rm{kg}}} \right)}}} \\\\ = 17.399842 \times {10^4}\\\\ = 173998.42\\\\ = 174 \times {10^3}\;{\rm{m/s}}\\\end{array}$

(b)

The expression for the conservation of energy from equation (4) is,

$\frac{1}{2}{m_e}v_f^2 - \frac{1}{2}{m_e}v_i^2 = {q_e}\left( {\Delta V} \right)$

Here,${m_e}$is the mass of the electron and${q_e}$is the charge of the electron.

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

$\begin{array}{c}\\\frac{1}{2}{m_e}v_f^2 - \frac{1}{2}{m_e}\left( {0\;{\rm{m/s}}} \right) = {q_e}V\\\\\frac{1}{2}{m_e}v_f^2 = {q_e}V\\\end{array}$

Rearrange the above equation for${v_f}$.

The expression for the speed of the electron that is accelerated from rest is,

$\frac{1}{2}{m_e}v_f^2 = {q_e}V$

Rearrange the above equation for v.

${v_f} = \sqrt {\frac{{2{q_e}V}}{{{m_e}}}}$

Substitute $\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)$for${q_e}$, 158 V for$\Delta V$, and $\left( {9.1 \times {{10}^{ - 31}}\;{\rm{kg}}} \right)$for${m_e}$.

$\begin{array}{c}\\{v_f} = \sqrt {\frac{{2\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {158\;{\rm{V}}} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}\;{\rm{kg}}} \right)}}} \\\\ = 7.45388755 \times {10^6}\\\\ = 745 \times {10^4}\;{\rm{m/s}}\\\end{array}$

Ans: Part a

The speed of the protons that are accelerated from rest through the potential difference is$174 \times {10^3}\;{\rm{m/s}}$.

Part b

The speed of the electrons that are accelerated from rest through the potential difference is$745 \times {10^4}\;{\rm{m/s}}$.