An electron with a speed of 1.3 × 107 m/s moves horizontally into a region where a constant vertical force of 4.4 × 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 29 mm horizontally.
2.4mm is a wrong answear
To determine the vertical distance the electron is deflected, we need to consider the horizontal and vertical motions separately.
First, let's calculate the time it takes for the electron to move 29 mm horizontally. We can use the formula:
time = distance / velocity
Given:
Distance = 29 mm = 0.029 m
Velocity = 1.3 × 10^7 m/s
Plugging in the values, we get:
time = 0.029 m / (1.3 × 10^7 m/s) ≈ 2.231 × 10^-9 seconds
Now, let's focus on the vertical motion. We know the force acting on the electron and its mass, so we can use Newton's second law of motion to find the vertical acceleration:
force = mass × acceleration
Given:
Force = 4.4 × 10^-16 N
Mass = 9.11 × 10^-31 kg
Plugging in the values, we get:
4.4 × 10^-16 N = (9.11 × 10^-31 kg) × acceleration
Solving for acceleration:
acceleration = (4.4 × 10^-16 N) / (9.11 × 10^-31 kg) ≈ 4.82 × 10^14 m/s^2
Now that we know the acceleration, we can use the kinematic equation to find the vertical displacement:
displacement = initial velocity × time + (1/2) × acceleration × time^2
Given:
Initial velocity = 0 m/s (since the electron starts from rest vertically)
Time = 2.231 × 10^-9 seconds
Acceleration = 4.82 × 10^14 m/s^2
Plugging in the values, we get:
displacement = 0 × (2.231 × 10^-9) + (1/2) × (4.82 × 10^14) × (2.231 × 10^-9)^2 ≈ 4.85 × 10^-5 meters
Therefore, the vertical distance the electron is deflected during the time it has moved 29 mm horizontally is approximately 4.85 × 10^-5 meters or 48.5 micrometers.