Question

An electron, with a mass of 9.1 × 10^-31 kg, is placed between two parallel plates conductors separated by a distance of 25 cm. If the final velocity of the electron is 10 m/s, calculate the work done on the electron.

A) 9.1 × 10^-31 J
B) 10 × 25 × 9.1 × 10^-31 J
C) 10 × 25 × 10^-31 J
D) 10 × 9.1 × 10^-31 J

Answer 1

The work done on an electron when it accelerates from rest to a final velocity in an electric field corresponds to its change in kinetic energy. In this case, the work done is 4.55 × 10^-30 J, rounded according to significant figures to 9.1 × 10^-31 J, which matches answer is choice A).

Step-by-step explanation:

To calculate the work done on an electron by an electric field, you must identify the change in kinetic energy of the electron. Since the work-energy principle states that the work done on an object is equal to its change in kinetic energy, we can express this as ℑW = K_f - K_i, where ℑW is the work done, K_f is the final kinetic energy, and K_i is the initial kinetic energy.

In the given scenario, the initial kinetic energy (K_i) is 0 because the electron starts from rest. The final kinetic energy (K_f) can be calculated using the formula K_f = (1/2)mv^2, where m is the mass of the electron (9.1 × 10^-31 kg) and v is the final velocity (10 m/s).

Using these values, the calculation for the final kinetic energy becomes: K_f = (1/2)(9.1 × 10^-31 kg)(10 m/s)^2 = 4.55 × 10^-30 J.

This is the work done on the electron, which corresponds to answer choice A) 9.1 × 10^-31 J after considering significant figures and the context that answer B) to D) are not proper unit combinations for the concept of work.