Question

An electron is accelerated through a potential difference of 0.20 v. How much greater would its final speed be if it is accelerated with four times as much voltage?

Answer 1

To calculate the final speed of an electron accelerated through a potential difference, you can use the formula v = sqrt(2qV/m), where q is the charge of the electron, V is the potential difference, and m is the mass of the electron. By plugging in the values given and solving the equation, you can determine the final speed for a potential difference of 0.20 V. To find the increase in final speed with four times the voltage, you can calculate the new final speed using the new potential difference and compare it to the original final speed.

Step-by-step explanation:

In physics, the speed of an electron can be determined using the formula:

v = sqrt(2qV/m)

Where:

v is the final speed of the electron

q is the charge of the electron (-1.60 × 10-19 C)

V is the potential difference

m is the mass of the electron (9.11 × 10-31 kg)

If an electron is accelerated through a potential difference of 0.20 V, its final speed can be calculated as:

v = sqrt(2 * (-1.60 × 10-19) * 0.20 / (9.11 × 10-31))

Using this formula, the final speed can be determined. To find out how much greater the final speed would be if the electron is accelerated with four times as much voltage, you can plug in the new potential difference (0.20 V * 4) and calculate the new final speed. The difference between the new final speed and the original final speed will give you the answer.

Answer 2

v₂ = 2* v₁

Step-by-step explanation:

• Assuming no friction present, the change in the electrostatic potential energy of the electron, must be equal in magnitude to the change in the kinetic energy:

`\Delta U_(ep) = (-e)*\Delta V= - \Delta K = -(1)/(2)*m*v^(2)`

• We know that ΔV = 0.2 V, so we can write the following equality:

`e* 0.2V = (1)/(2) * m *v_(1) ^(2) (1)`

• Now, if the voltage increases 4 times, we can write the following equality:

`e* 0.8V = (1)/(2) * m *v_(2) ^(2) (2)`

• Dividing both sides in (1) and (2), and simplifying common terms, we get:

`(v_(2) ^(2))/(v_(1) ^(2)) = 4\\\\ v_(2) ^(2) = 4*v_(1) ^(2)`

• Taking square roots at both sides, we get:
• v₂ = 2* v₁
• The final speed, if the electron is accelerated through a potential difference four times larger, will be double than the one for a potential difference of 0.2 V.