Chapter 2, Problem 52P

(a)

To determine

The speed of the electrons when they hit the screen.

Answer to Problem 52P

The speed of the electrons is $1.7\times {10}^6\text{m}/\text{s}$.

Explanation of Solution

Write the equation of motion according to the Newton’s second law.

$v_{fx}^2-v_{ix}^2=2a_x\Delta x$

Here, $v_{fx}$ is the final velocity, $v_{ix}$ is the initial velocity along $x$ direction, $a_x$ is the acceleration along $x$ direction and $\Delta x$ is the change in distance.

Here, the initial velocity is $0\text{m}/\text{s}$.

$\begin{array}{l}{v}_{fx}^2=2a\Delta x\\ v_{fx}=\sqrt{2a\Delta x}\end{array}$ (I)

Conclusion:

Substitute $7.03\times {10}^{13}\text{m}/{\text{s}}^2$ for $a_x$ and $2.0\text{cm}$ for $\Delta x$ in the equation I to find $v_{fx}$.

$\begin{array}{c}{v}_{fx}=\sqrt{2\left(7.03\times {10}^{13}\text{m}/{\text{s}}^2\right)\left(2.0\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}\\ =1.7\times {10}^6\text{m}/\text{s}\end{array}$

Therefore, the speed of the electrons is $1.7\times {10}^6\text{m}/\text{s}$.

(b)

To determine

The time taken for the electrons to travel the length of the tube.

Answer to Problem 52P

The time taken for the electrons to travel the length of the tube is $290\text{ns}$

Explanation of Solution

Write the expression for the equation of motion for constant $a_x$.

$\Delta x=v_{ix}\Delta t+\frac{1}{2}{a}_x{\left(\Delta t\right)}^2$

Here, $\Delta t$ is the change in time.

For the $x$ from 0 to $2.0\text{cm}$,

$\begin{array}{l}\Delta x_1=\frac{1}{2}{a}_x\Delta t_1^2\\ \Delta t_1^2=\frac{2\Delta x_1}{a_x}\\ \Delta t_1=\sqrt{\frac{2\Delta x_1}{a_x}}\end{array}$ (II)

For the $x$ from $2.0\text{cm}$ to $47\text{cm}$,

$\begin{array}{l}\Delta x_2=v_{fx}\Delta t_2\\ \Delta t_2=\frac{\Delta x_2}{v_{fx}}\end{array}$ (III)

Write the expression for the total time.

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

Substitute equation II and III in equation IV,

$\Delta t=\sqrt{\frac{2\Delta x_1}{a_x}}+\frac{\Delta x_2}{v_{fx}}$ (V)

Conclusion:

Substitute $2.0\text{cm}$ for $\Delta x_1$, $7.03\times {10}^{13}\text{m}/{\text{s}}^2$ for $a_x$, $45\text{cm}$ for $\Delta x_2$ and $1.7\times {10}^6\text{m}/\text{s}$ for $v_{fx}$ in equation V to find $\Delta t$.

$\begin{array}{c}\Delta t=\sqrt{\frac{2\left(2.0\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}{7.03\times {10}^{13}\text{m}/{\text{s}}^2}}+\frac{\left(45\text{cm}\times \frac{1\text{m}}{{10}^2\text{cm}}\right)}{1.677\times {10}^6\text{m}/\text{s}}\\ =290\text{ns}\end{array}$

Therefore, the time taken for the electrons to travel the length of the tube is $290\text{ns}$