Chapter 26, Problem 99P

(a)

To determine

The classically predicted speed of electron.

Answer to Problem 99P

The classically predicted speed of electron is $\underset{\_}{2.965\times {10}^9\text{m}/\text{s}}$.

Explanation of Solution

Given that the accelerating potential is $25.00\text{MV}$.

Write the classical expression for the kinetic energy of the electron.

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

Here, $K$ is the kinetic energy of the electron, $m$ is the mass of the electron, $v$ is the speed  of the electron.

Write the expression for the kinetic energy of the electron in terms of the accelerating potential.

  $K=e\Delta V$                                                                                                                 (II)

Here, $e$ is the charge of the electron, $\Delta V$ is the accelerating potential.

Equate the right-hand sides of equation (I) and (II) and solve for $v$.

    $\begin{array}{c}\frac{1}{2}m{v}^2=e\Delta V\\ v=\sqrt{\frac{2e\Delta V}{m}}\end{array}$                                                                                                   (III)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $25.00\text{MV}$ for $\Delta V$, and $9.109\times {10}^{-31}\text{kg}$ for $m$ in equation (III) to find $v$.

  $\begin{array}{c}v=\sqrt{\frac{2\left(1.602\times {10}^{-19}\text{C}\right)\left(25.00\times {10}^6\text{V}\right)}{9.109\times {10}^{-31}\text{kg}}}\\ =\sqrt{\frac{2\left(1.602\times {10}^{-19}\text{C}\right)\left(25.00\text{MV}\times \frac{1\times {10}^6\text{V}}{1\text{MV}}\right)}{9.109\times {10}^{-31}\text{kg}}}\\ =2.965\times {10}^9\text{m}/\text{s}\end{array}$

Therefore, The classically predicted speed of electron is $\underset{\_}{2.965\times {10}^9\text{m}/\text{s}}$.

(b)

To determine

The actual speed of the electron.

Answer to Problem 99P

The actual speed of the electron is $\underset{\_}{2.99\times {10}^8\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the relativistic kinetic energy.

    $K=\left(\gamma -1\right)m{c}^2$                                                                                                       (IV)

Here, $K$ is the relativistic kinetic energy, $\gamma$ is the Lorentz factor, $m$ is the mass of the electron, $c$ is the speed of the light.

Write the expression for the Lorentz factor.

    $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$                                                                                                        (V)

Here, $v$ is the speed of electron, $c$ is the speed of light.

Use equation (V) in equation (IV).

    $K=\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)m{c}^2$                                                                                      (VI)

Use equation (II) in (VI) and solve for $v$.

    $\begin{array}{c}e\Delta V=\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)m{c}^2\\ {\left(1+\frac{e\Delta V}{m{c}^2}\right)}^2=\frac{1}{1-v^2/c^2}\\ 1-\frac{v^2}{c^2}={\left(1+\frac{e\Delta V}{m{c}^2}\right)}^{-2}\\ v=c\sqrt{1-{\left(1+\frac{e\Delta V}{m{c}^2}\right)}^{-2}}\end{array}$                                                                      (VII)

Conclusion:

Substitute $2.99\times {10}^8\text{m}/\text{s}$ for $c$ , $1.602\times {10}^{-19}\text{C}$ for $e$ , $25.00\text{MV}$ for $\Delta V$, $9.109\times {10}^{-31}\text{kg}$ for $m$ in equation (VI) to find $v$.

    $\begin{array}{c}v=\left(2.99\times {10}^8\text{m}/\text{s}\right)\sqrt{1-{\left(1+\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(25.00\text{MV}\right)}{\left(9.109\times {10}^{-31}\text{kg}\right){\left(2.998\times {10}^8\text{m}/\text{s}\right)}^2}\right)}^{-2}}\\ =\left(2.99\times {10}^8\text{m}/\text{s}\right)\sqrt{1-{\left(1+\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(25.00\text{MV}\times \frac{{10}^6\text{V}}{1\text{MV}}\right)}{\left(9.109\times {10}^{-31}\text{kg}\right){\left(2.998\times {10}^8\text{m}/\text{s}\right)}^2}\right)}^{-2}}\\ =2.99740\times {10}^8\text{m}/\text{s}\end{array}$

Therefore, the actual speed of the electron is $\underset{\_}{2.99\times {10}^8\text{m}/\text{s}}$.