Question

An electron is to be accelerated from a velocity of 2.50×10⁶ m/s to a velocity of 7.50×10⁶ m/s. Through what potential difference must the electron pass to accomplish this?

Answer 1

To find the potential difference required to accelerate an electron from a velocity of 2.50×10¶ m/s to 7.50×10¶ m/s, one should calculate the change in kinetic energy and use the electron's charge to compute the voltage.

Step-by-step explanation:

To determine through what potential difference an electron must be accelerated to increase its velocity from 2.50×10¶ m/s to 7.50×10¶ m/s, we can use the concept of conservation of energy. The work done by the electric field on the electron equals the change in its kinetic energy.

First, we calculate the initial and final kinetic energies (KE) using KE = 0.5 × mass × (velocity)^2, where the mass of the electron (m_e) is 9.11×10^-31 kg.

Initial kinetic energy (KE_initial) = 0.5 × m_e × (2.50×10¶ m/s)^2

Final kinetic energy (KE_final) = 0.5 × m_e × (7.50×10¶ m/s)^2

The change in kinetic energy (KE_change) = KE_final - KE_initial.

The work done by the electric field (Work) is equal to the change in kinetic energy and also equal to the charge (q_e) on the electron multiplied by the potential difference (V): Work = q_e × V = KE_change.

Since the charge of an electron (q_e) is -1.60×10^-19 C, we can solve for V: V = KE_change / q_e.

After calculating KE_change and dividing by the charge of the electron, we will get the potential difference the electron must be accelerated through.