In a cathode ray tube, electrons initially at rest are accelerated by a uniform electric field of magnitude 4.71 x 105 N/C during the first 5.49 cm of the tube's length; then they move at essentially constant velocity another 47.9 cm before hitting the screen. Find the speed of the electrons when they hit the screen.
How long does it take them to travel the length of the tube?
To find the speed of the electrons when they hit the screen, we need to use the acceleration-distance equation. The initial speed of the electrons is zero, so the equation can be simplified to:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity
- vi is the initial velocity (which is zero in this case)
- a is the acceleration
- d is the distance
First, let's find the acceleration using the given electric field magnitude:
a = qE/m
Where:
- q is the charge of the electron (-1.6 x 10^-19 C)
- E is the electric field magnitude (4.71 x 10^5 N/C)
- m is the mass of the electron (9.11 x 10^-31 kg)
Plugging in the values, we get:
a = (-1.6 x 10^-19 C) * (4.71 x 10^5 N/C) / (9.11 x 10^-31 kg) ≈ -8.31 x 10^13 m/s^2
Since the electric field accelerates the electrons, the acceleration is negative.
Next, let's calculate the distance covered during the first 5.49 cm:
d = 5.49 cm = 5.49 x 10^-2 m
Now, plug the values into the acceleration-distance equation:
vf^2 = 0 + 2 * (-8.31 x 10^13 m/s^2) * (5.49 x 10^-2 m)
vf^2 ≈ -9.12 x 10^12 m^2/s^2
The velocity (vf) cannot be negative, so we take the magnitude of vf:
vf = √[(-9.12 x 10^12 m^2/s^2)]
vf ≈ 3.02 x 10^6 m/s
Therefore, the speed of the electrons when they hit the screen is approximately 3.02 x 10^6 m/s.
To find the time taken to travel the length of the tube, we need to know their constant velocity. We can calculate this using the distance and the speed previously found.
Let's consider the remaining distance:
d_remaining = Total distance - Distance covered during first 5.49 cm
d_remaining = 47.9 cm - 5.49 cm = 42.41 cm = 0.4241 m
Since the electrons are moving at a constant velocity, we can use the equation:
v = d/t
Where:
- v is the constant velocity
- d is the distance
- t is the time taken
Rearranging the equation to solve for time, we get:
t = d/v
Plugging in the values, we have:
t = (0.4241 m) / (3.02 x 10^6 m/s)
t ≈ 1.40 x 10^-7 s
Therefore, it takes approximately 1.40 x 10^-7 seconds for the electrons to travel the length of the tube.
To find the speed of the electrons when they hit the screen, we can use the concept of electric potential energy.
Given:
Electric field magnitude (E) = 4.71 x 10^5 N/C
Distance of acceleration (d1) = 5.49 cm = 0.0549 m
Distance of constant velocity (d2) = 47.9 cm = 0.479 m
1. Let's calculate the work done on an electron by the electric field during the acceleration phase:
Work (W) = force (F) x distance (d1)
W = e * E * d1
Where e is the charge of an electron (1.6 x 10^-19 C)
2. The work done on the electron is converted into its kinetic energy:
KE = W
KE = 0.5 * m * v^2
Where m is the mass of the electron (9.1 x 10^-31 kg) and v is the velocity of the electron.
3. Set the equations for work and kinetic energy equal to each other:
e * E * d1 = 0.5 * m * v^2
4. Solve for v:
v^2 = (2 * e * E * d1) / m
v = sqrt((2 * e * E * d1) / m)
5. Now, let's calculate the time taken to travel the entire length of the tube:
Total distance traveled (d_total) = d1 + d2
6. The time taken (t_total) is equal to the total distance divided by the final velocity:
t_total = d_total / v
7. Plug in the given values and calculate the final velocity and total time.
Note: Remember to convert all the units to SI units before performing the calculations.
Let's solve it step by step:
1. Calculate work done on the electron during acceleration:
W = (1.6 x 10^-19 C) * (4.71 x 10^5 N/C) * 0.0549 m
2. Calculate the kinetic energy:
KE = W
3. Solve for v:
v = sqrt((2 * (1.6 x 10^-19 C) * (4.71 x 10^5 N/C) * 0.0549 m) / (9.1 x 10^-31 kg))
4. Calculate the total distance traveled:
d_total = 0.0549 m + 0.479 m
5. Calculate the total time taken:
t_total = (0.0549 m + 0.479 m) / v
Plug in the values and calculate the answer.