Chapter 2, Problem 2.1HP

A free electron has an initial potential energy perunit charge (voltage) of 17 kJ/C and a velocity of93 Mm/s. Later, it potential energy per unit chargeis 6 kJ/C. Determine the change in velocity of the electron.

To determine

The change in velocity of the electron.

Answer to Problem 2.1HP

The change in velocity of the electron is $23.83\text{Mm}/\text{s}$ .

Explanation of Solution

Calculation:

The expression for the difference in the potential energy is given by,

  $\Delta P.E=Q_e\left(V_f-V_i\right)$

Here, $Q_e$ is the charge on an electron.

The expression for the difference in the kinetic energy is given by,

  $\Delta K.E=\frac{1}{2}{m}_e\left(U_f^2-U_i^2\right)$

Here, $m_e$ is the mass of an electron.

The expression for the first law of thermodynamics is given by,

  $Q_{\text{heat}}-W=\Delta K.E+\Delta P.E+\Delta P.E_g$

Here, $\Delta P.E_g$ is the gravitational potential energy which is equal to $0$ and $W$ is the work done.

There is no external force is applied, so the value of the work done is $0$ .

Substitute $0$ for $Q_{\text{heat}}$, $0$ for $W$, $\frac{1}{2}{m}_e\left(U_f^2-U_i^2\right)$ for $\Delta K.E$, $Q_e\left(V_f-V_i\right)$ for $\Delta P.E$ and $0$ for $\Delta P.E_g$ in the above equation.

  $\begin{array}{l}0-0=\frac{1}{2}{m}_e\left(U_f^2-U_i^2\right)+Q_e\left(V_f-V_i\right)+0\\ \left(U_f^2-U_i^2\right)=-\frac{2Q_e}{m_e}\left(V_f-V_i\right)\\ U_f^2=U_i^2-\frac{2Q_e}{m_e}\left(V_f-V_i\right)\end{array}$

Substitute $93\text{Mm}/\text{s}$ for $U_i$, $1.6\times {10}^{-19}\text{C}$ for $Q_e$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $6\text{kJ}/\text{C}$ for $V_f$ and $17\text{kJ}/\text{C}$ for $V_i$ in the above equation.

  $\begin{array}{c}{U}_f^2={\left(93\text{Mm}/\text{s}\right)}^2-\frac{2\left(1.6\times {10}^{-19}\right)}{9.11\times {10}^{-31}}\left(6\text{kJ}/\text{C}-17\text{kJ}/\text{C}\right)\\ ={\left(93\text{Mm}/\text{s}\right)}^2-\frac{2\left(1.6\times {10}^{-19}\right)}{9.11\times {10}^{-31}}\left(-11\text{kJ}/\text{C}\right)\end{array}$

The conversion of $\text{Mm}/\text{s}$ into $\text{m}/\text{s}$ is given by,

  $\text{1}\text{Mm}/\text{s}={10}^6\text{m}/\text{s}$

The conversion of $\text{93}\text{Mm}/\text{s}$ into $\text{m}/\text{s}$ is given by,

  $\text{93}\text{Mm}/\text{s}=93\times {10}^6\text{m}/\text{s}$

The conversion of $\text{kJ}/\text{C}$ into $\text{J}/\text{C}$ is given by,

  $\text{1}\text{kJ}/\text{C}={10}^3\text{kJ}/\text{C}$

The conversion of $11\text{kJ}/\text{C}$ into $\text{J}/\text{C}$ is given by,

  $11\text{kJ}/\text{C}=11\times {10}^3\text{J}/\text{C}$

Rewrite the above equation for the final velocity.

  $\begin{array}{c}{U}_f^2={\left(93\times {10}^6\right)}^2-\frac{2\left(1.6\times {10}^{-19}\right)}{9.11\times {10}^{-31}}\left(-11\times {10}^3\right)\\ =4.785\times {10}^{15}\\ U_f=\sqrt{4.785\times {10}^{15}}\\ =6.917\times {10}^7\text{m}/\text{s}\end{array}$

The conversion of $\text{m}/\text{s}$ into $\text{Mm}/\text{s}$ is given by,

  $\text{1}\text{m}/\text{s}={10}^{-6}\text{Mm}/\text{s}$

The conversion of $6.917\times {10}^7\text{m}/\text{s}$ into $\text{Mm}/\text{s}$ is given by,

  $\begin{array}{c}6.917\times {10}^7\text{m}/\text{s}=6.917\times {10}^7\times {10}^{-6}\text{Mm}/\text{s}\\ =69.17\text{Mm}/\text{s}\end{array}$

The change in the velocity of the electron is given by,

  $\Delta U=U_f-U_i$

Substitute $69.17\text{Mm}/\text{s}$ for $U_f$ and $93\text{Mm}/\text{s}$ for $U_i$ in the above equation.

  $\begin{array}{c}\Delta U=69.17-93\\ =-23.83\text{Mm}/\text{s}\\ \left|\Delta U\right|=23.83\text{Mm}/\text{s}\end{array}$

Conclusion:

Therefore,the change in velocity of the electron is $23.83\text{Mm}/\text{s}$ .