91Ó°ÊÓ

• Speed of electron, $1.2×{10}^7\mathrm{m}/\mathrm{s}$,
• The magnitude of constant vertical force$4.5×{10}^{-16}\mathrm{N}$
• Mass of electron$9.1×{10}^{-31}\mathrm{kg}$
• Horizontal distance travel is$30×{10}^{-3}\mathrm{m}$

The force on the electron is equal to the product of mass and acceleration of the electron. We can say that acceleration of the electron is vertical and for all practical purposes, the only force acting on it is the electric force.

We take the positive X direction of initial velocity and the positive Y direction to be the direction of electric force.

Formulae:

$\mathrm{F}=\mathrm{ma}$ (1)

$\mathrm{s}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (2)

Here, F is force, is the mass of an electron, is acceleration, S is displacement, $V_0$is the initial velocity, and t is time.

To calculate the acceleration, use equation (1) and substitute the value of mass and force on the electron,

$$\begin{gathered} \mathrm{F}=\mathrm{ma} \\ \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \\ =\frac{4.5×{10}^{-16}\mathrm{N}}{9.1×{10}^{-31}\mathrm{kg}} \\ =4.94×{10}^{14}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the electron is$4.94×{10}^{14}\mathrm{m}/{\mathrm{s}}^2$

There is no acceleration in the horizontal direction. Therefore, using equation (ii) and the horizontal displacement, we can find the time. Substitute the values in equation (ii).

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_0\mathrm{t} \\ 30×{10}^{-3}\mathrm{m}=\left(1.2×{10}^7\mathrm{m}/\mathrm{s}\right)\mathrm{t} \\ \mathrm{t}=\frac{30×{10}^{-3}\mathrm{m}}{1.2×{10}^7\mathrm{m}/\mathrm{s}} \\ \mathrm{t}=2.5×{10}^{-9}\mathrm{s} \end{gathered}$$

Therefore, the time to travel the horizontal distance is$2.5×{10}^{-9}s$.

The initial velocity in the vertical direction is zero. Therefore, using equation (ii) again, we can find the vertical displacement. Substitute the value of acceleration and time found above.

$$\begin{gathered} \mathrm{y}=\frac{1}{2}{\mathrm{at}}^2 \\ =\frac{1}{2}\left(4.94×{10}^{14}\mathrm{m}/{\mathrm{s}}^2\right){\left(2.5×{10}^{-9}\mathrm{s}\right)}^2 \\ =1.5×{10}^{-3}\mathrm{m} \end{gathered}$$

Therefore, the vertical distance traveled by the electron is$1.5×{10}^{-3}\mathrm{m}$.