Question

An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 19 mm horizontally.

Answer 1

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

Answer 2

Answer:

0.85mm

Explanation:

It is given that,

Speed of the electron in horizontal region, $v=1.2\times 10^7\ m/s$

Vertical force, $F_y=5.2\times 10^{-16}\ N$

Vertical acceleration, $a_y=\dfrac{F_y}{m}$

$a_y=\dfrac{5.2\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg} \\a_y=5.708\times 10^{14}\ m/s^2..........(1)$

Let t is the time taken by the electron, such that,

$t=\dfrac{x}{v_x}\\t=\dfrac{19\times 10^-^3 \ m}{1.1\times 10^7\ m/s}\\t=1.727\times 10^{-9}\ s...........(2)$

Let d_y is the vertical distance deflected during this time. It can be calculated using second equation of motion:

$d_y=ut+\dfrac{1}{2}a_yt^2$

u = 0

$d_y=\dfrac{1}{2}\times 5.708\times 10^{14}\ m/s^2\times (1.727\times 10^{-9}\ s)^2$

$d_y=0.0008512\ m\\d_y=0.85\ mm$