Chapter 4, Problem 4P

It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope, however, can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The “lenses” of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity ${\stackrel{\to }{v}}_i=v_i\hat{i}$. As it passes through the region x = 0 to x = d, the electron experiences acceleration $\stackrel{\to }{a}(t)=a_x\hat{i}+a_y\hat{j}$, where a_x and a_y are constants. For the case v_i = 1.80 × 10⁷ m/s, a_x = 8.00 × 10¹⁴ m/s², and a_y = 1.60 × 10¹⁵ m/s² determine at x = d = 0.010 0 m (a) the position of the electron, (b) the velocity of the electron, (c) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis).

To determine

The electron position after travelling a horizontal distance of $0.0100\text{m}$ .

Answer to Problem 4P

The position of the electron is $\left(0.010\text{m}\right)\widehat{i}+\left(2.41\times {10}^{-4}\text{ m}\right)\widehat{j}$.

Explanation of Solution

The x component of velocity is $1.80\times {10}^7\text{m}/\text{s}$, the x component of the acceleration is $8.00\times {10}^{14}\text{m}/{\text{s}}^2$, the y component of acceleration is $1.60\times {10}^{15}\text{m}/{\text{s}}^2$ and the final horizontal distance is $0.0100\text{ m}$.

Write the equation for the x component of the motion

    $x_{\text{f}}=x_{\text{i}}+v_{\text{x}}t+\frac{1}{2}{a}_{\text{x}}{t}^2$

Here, $x_{\text{f}}$ is the final distance, $x_{\text{i}}$ is the initial distance, $v_{\text{x}}$ is the x component of velocity and $a_{\text{x}}$ is the x component of acceleration.

Substitute $0.0100\text{ m}$ for $x_{\text{f}}$, $0$ for $x_{\text{i}}$, $1.80\times {10}^7\text{m}/\text{s}$ for $v_{\text{x}}$ and $8.00\times {10}^{14}\text{m}/{\text{s}}^2$ for $a_{\text{x}}$ in above equation.

$\begin{array}{c}\left(0.0100\text{ m}\right)=\left[\begin{array}{l}0+\left(1.80\times {10}^7\text{m}/\text{s}\right)t\\ +\frac{1}{2}\left(8.00\times {10}^{14}\text{m}/{\text{s}}^2\right)t^2\end{array}\right]\\ \left(4.00\times {10}^{14}\text{m}/{\text{s}}^2\right)t^2+\left(1.80\times {10}^7\text{m}/\text{s}\right)t-0.01\text{ m}=\text{0}\\ t=5.49\times {10}^{-10}\text{ s}\end{array}$

Write the equation for the y component of the motion

    $y_{\text{f}}=y_{\text{i}}+v_{\text{y}}t+\frac{1}{2}{a}_{\text{y}}{t}^2$

Here, $y_{\text{f}}$ is the final distance, $y_{\text{i}}$ is the initial distance, $v_{\text{y}}$ is the y component of velocity and $a_{\text{y}}$ is the y component of acceleration.

Conclusion:

Substitute $0$ for $y_{\text{i}}$, $0$ for $v_{\text{y}}$, $5.49\times {10}^{-10}\text{ s}$ for $t$ and $1.60\times {10}^{15}\text{m}/{\text{s}}^2$ for $a_{\text{y}}$ in above equation to find $y_{\text{f}}$.

    $\begin{array}{c}{y}_{\text{f}}=0+\left(0\right)\left(5.49\times {10}^{-10}\text{ s}\right)+\frac{1}{2}\left(1.60\times {10}^{15}\text{m}/{\text{s}}^2\right){\left(5.49\times {10}^{-10}\text{ s}\right)}^2\\ =\frac{1}{2}\left(1.60\times {10}^{15}\text{m}/{\text{s}}^2\right){\left(5.49\times {10}^{-10}\text{ s}\right)}^2\\ =2.41\times {10}^{-4}\text{ m}\end{array}$

Therefore, the position of the electron is $\left(0.010\text{m}\right)\widehat{i}+\left(2.41\times {10}^{-4}\text{ m}\right)\widehat{j}$.

To determine

The electron velocity after travelling a horizontal distance of $0.0100\text{m}$ .

Answer to Problem 4P

The velocity of the electron is $\left(1.84\times {10}^7\text{m}/{\text{s}}^2\right)\widehat{i}+\left(8.78\times {10}^5\text{m}/\text{s}\right)\widehat{j}$.

Explanation of Solution

Write the formula to calculate velocity

    $\begin{array}{c}{v}_{\text{f}}=v_{\text{i}}+at\\ =\left(v_{\text{x}}+a_{\text{x}}t\right)\widehat{i}+\left(\left(v_{\text{y}}+a_{\text{y}}t\right)\right)\widehat{j}\end{array}$

Conclusion:

Substitute $1.80\times {10}^7\text{m}/\text{s}$ for $v_{\text{x}}$, $0$ for $v_{\text{y}}$, $8.00\times {10}^{14}\text{m}/{\text{s}}^2$ for $a_{\text{x}}$, $5.49\times {10}^{-10}\text{ s}$ for $t$ and $1.60\times {10}^{15}\text{m}/{\text{s}}^2$ for $a_{\text{y}}$ in above equation to find $v_{\text{f}}$.

    $\begin{array}{c}{v}_{\text{f}}=\left[\begin{array}{l}\left(1.80\times {10}^7\text{m}/\text{s}+\left(8.00\times {10}^{14}\text{m}/{\text{s}}^2\right)\left(5.48\times {10}^{-10}\text{ s}\right)\right)\widehat{i}\\ +\left(0+\left(1.60\times {10}^{15}\text{m}/{\text{s}}^2\right)\left(5.49\times {10}^{-10}\text{ s}\right)\right)\widehat{j}\end{array}\right]\\ =\left(1.84\times {10}^7\text{m}/{\text{s}}^2\right)\widehat{i}+\left(8.78\times {10}^5\text{m}/\text{s}\right)\widehat{j}\end{array}$

Therefore, the velocity of the electron is $\left(1.84\times {10}^7\text{m}/{\text{s}}^2\right)\widehat{i}+\left(8.78\times {10}^5\text{m}/\text{s}\right)\widehat{j}$.

To determine

The electron speed after travelling a horizontal distance of $0.0100\text{m}$.

Answer to Problem 4P

The speed of the electron is $1.8\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

Write the formula to calculate speed

    $\begin{array}{c}v=\left|\overrightarrow{v}\right|\\ =\sqrt{v_{\text{x}}^2+v_{\text{y}}^2}\end{array}$

Substitute $\left(1.84\times {10}^7\text{m}/{\text{s}}^2\right)$ for $v_{\text{x}}$ and $\left(8.78\times {10}^5\text{m}/\text{s}\right)$ for $v_{\text{y}}$ to find the $v$.

    $\begin{array}{c}v=\sqrt{{\left(1.84\times {10}^7\text{m}/{\text{s}}^2\right)}^2+{\left(8.76\times {10}^5\text{m}/\text{s}\right)}^2}\\ =1.85\times {10}^7\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the electron is $1.8\times {10}^7\text{m}/\text{s}$.

To determine

The direction of motion of the electron.

Answer to Problem 4P

The direction of travel of the electron is $2.73°$ from x axis.

Explanation of Solution

Write the formula to calculate direction of travel

    $\theta ={\tan }^{-1}\left(\frac{v_{\text{yf}}}{v_{\text{xf}}}\right)$

Conclusion:

Substitute $1.85\times {10}^7\text{m}/{\text{s}}^2$ for $v_{\text{xf}}$ and $\left(8.78\times {10}^5\text{m}/\text{s}\right)$ for $v_{\text{yf}}$ to find the direction.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{8.78\times {10}^5\text{m}/\text{s}}{1.84\times {10}^7\text{m}/{\text{s}}^2}\right)\\ =2.73°\end{array}$

Therefore, the direction of travel of the electron is $2.73°$ from x axis.