Chapter 16, Problem 101P

(a)

To determine

The speed of electrons when they hit the screen.

Answer to Problem 101P

The speed of electrons when they hit the screen is $\underset{\_}{8.4\times {10}^7\text{m/s}}$.

Explanation of Solution

Write the expression for Newton’s equation of motion.

$v_{\text{f}}^2-v_{\text{i}}^2=2a\Delta x$ (I)

Here, $v_{\text{f}}$ is the final velocity, $v_{\text{i}}$ is the initial velocity, $a$ is the acceleration, and $\Delta x$ is the distance travelled by 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.

Rewrite equation (I) to find $v_{\text{f}}$.

$\begin{array}{l}{v}_{\text{f}}^2-0=2a\Delta x\\ v_{\text{f}}=\sqrt{2\left(\frac{eE}{m}\right)\Delta x}\end{array}$ (III)

Conclusion:

Substitute, $1.602\times {10}^{-19}\text{C}$ for $e$, $4.0\times {10}^2\text{N/C}$ for $E$, $9.109\times {10}^{-31}\text{kg}$ for $m$, $0.050\text{m}$ for $\Delta x$ in equation (III).

$\begin{array}{c}{v}_{\text{f}}=\sqrt{2\left(\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(4.0\times {10}^5\text{N/C}\right)}{\left(9.109\times {10}^{-31}\text{kg}\right)}\right)\left(0.050\text{m}\right)}\\ =8.4\times {10}^7\text{m/s}\end{array}$

Therefore, the speed of electrons when they hit the screen is $\underset{\_}{8.4\times {10}^7\text{m/s}}$.

(b)

To determine

The time taken to travel the length of the tube.

Answer to Problem 101P

The time taken to travel the length of the tube is $\underset{\_}{6.6\text{ns}}$.

Explanation of Solution

Let $\Delta t_1$, and $\Delta t_2$ be the time intervals at which the acceleration is remains constant

Write the expression of Newton’s Laws of motion to find the displacement change.

$\begin{array}{c}\Delta x_1=v_{\text{i}}\Delta t_1+\frac{1}{2}a{\left(\Delta t_1\right)}^2\\ =v_{\text{i}}\Delta t_1+\frac{1}{2}\times \frac{eE}{M}{\left(\Delta t_1\right)}^2\end{array}$ (IV)

Substitute $0\text{m}/\text{s}$ for $v_{\text{i}}$ in the above equation and rewrite it to find $\Delta t_1$.

$\Delta t_1=\sqrt{\frac{2m\Delta x_1}{eE}}$ (V)

Write the expression for $\Delta t_2$.

$\begin{array}{c}\Delta x_2=v\Delta t_2\\ \Delta t_2=\frac{\Delta x_2}{v}\end{array}$ (VI)

Substitute equation (III) in (VI).

$\Delta t_2=\frac{\Delta x_2}{\sqrt{\frac{2eE\Delta x_1}{m}}}$ (VII)

Write the expression for total time interval.

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

Substitute equation (VI) and (V).

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

Conclusion:

Substitute $45\text{cm}$ for $\Delta x_2$, $5.0\text{cm}$ for $\Delta x_1$, $1.602\times {10}^{-19}\text{C}$ for $e$, $4.0\times {10}^2\text{N/C}$ for $E$, $9.109\times {10}^{-31}\text{kg}$ for $m$, $0.050\text{m}$ for $\Delta x_1$ in equation (IX).

$\begin{array}{c}\Delta t=\left(1+\frac{45\text{cm}}{2\left(5.0\text{cm}\right)}\right)\sqrt{\frac{2\left(9.109\times {10}^{-31}\text{kg}\right)\left(0.050\text{m}\right)}{\left(1.602\times {10}^{-19}\text{C}\right)\left(4.0\times {10}^2\text{N/C}\right)}}\\ =6.6\text{ns}\end{array}$

Therefore, the time taken to travel the length of the tube is $\underset{\_}{6.6\text{ns}}$.