Chapter 4, Problem 20P

(a)

To determine

Speed of electrons when they hit the screen.

Answer to Problem 20P

The speed of the electrons is $\underset{\_}{\text{8}\text{.84}\times {\text{10}}^{\text{7}}\text{m}/\text{s}}$

Explanation of Solution

Write the equation of motion.

    $v^2=u^2+2a\Delta x$        (I)

Here $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $\Delta x$ is the distance travelled by the electron.

Write the expression for acceleration from the balancing equation of forces

    $\begin{array}{c}ma=eE\\ a=\frac{eE}{m}\end{array}$        (II)

Here $e$ is the charge, $m$ is the mass and $E$ is the electric field.

Substitute $u=0$ and (II) in (I)

    $v=\sqrt{2\left(\frac{eE}{m}\right)\Delta x}$        (III)

Conclusion

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$, $4\times {10}^2\text{N}/\text{C}$ for $E$, $9\times {10}^{-31}\text{kg}$ for $m$ and $0.050\text{m}$ for $\Delta x$ in (III)

    $\begin{array}{c}v=\sqrt{2\left(\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(4\times {10}^2\text{N}/\text{C}\right)}{9\times {10}^{-31}\text{kg}}\right)\left(0.050\text{m}\right)}\\ =8.4\times {10}^7\text{m}/\text{s}\end{array}$

The speed of the electrons is $\underset{\_}{\text{8}\text{.84}\times {\text{10}}^{\text{7}}\text{m}/\text{s}}$

(b)

To determine

Time taken to travel the length of tube.

Answer to Problem 20P

The time taken to travel the length of tube is $\underset{\_}{\text{6}\text{.6}\text{ns}}$

Explanation of Solution

Write the equation of motion to find the displacement change.

    $\Delta x_1=u\Delta t_1+\frac{1}{2}a{\left(\Delta t_1\right)}^2$        (IV)

Substitute (II) in (IV)

    $\Delta x_1=u\Delta t_1+\frac{1}{2}\left(\frac{eE}{m}\right){\left(\Delta t_1\right)}^2$        (V)

Substitute $u=0$ in (V) and rearrange in terms of $\Delta t_i$

    $\Delta t_1=\sqrt{\frac{2\Delta x_{1}m}{eE}}$        (VI)

Write the equation for $\Delta t_2$

    $\Delta t_2=\frac{\Delta x_2}{v}$        (VII)

Substitute (III) in (VII)

    $\Delta t_2=\frac{\Delta x_2}{\sqrt{2\left(\frac{eE}{m}\right)\Delta x}}$        (VIII)

Write the equation for total time interval

    $\Delta t=\Delta t_1+\Delta t_2$        (IX)

Substitute (VI) and (VIII) in (IX)

    $\begin{array}{c}\Delta t=\sqrt{\frac{2\Delta x_{1}m}{eE}}+\frac{\Delta x_2}{\sqrt{2\left(\frac{eE}{m}\right)\Delta x}}\\ =\left(1+\frac{\Delta x_2}{2\Delta x_1}\right)\sqrt{\frac{2m\Delta x_1}{eE}}\end{array}$        (X)

Conclusion

Substitute $45\text{cm}$ for $\Delta x_2$, $5\text{cm}$ for $\Delta x_1$, $1.6\times {10}^{-19}\text{C}$ for $e$, $4\times {10}^2\text{N}/\text{C}$ for $E$, $9\times {10}^{-31}\text{kg}$ for $m$ and $0.050\text{m}$ for $\Delta x$ in (X)

    $\begin{array}{c}\Delta t=\left(1+\frac{45\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{2\left(5\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\right)\sqrt{\frac{2\left(9\times {10}^{-31}\text{kg}\right)\left(0.050\text{m}\right)}{\left(1.6\times {10}^{-19}\text{C}\right)\left(4\times {10}^2\text{N}/\text{C}\right)}}\\ =6.6\text{ns}\end{array}$

The time taken to travel the length of tube is $\underset{\_}{\text{6}\text{.6}\text{ns}}$