Final answer:
To find the electron's velocity after 57.1 ns in a 337 N/C electric field, calculate the force on the electron, determine the acceleration, and apply the kinematic equation v = at since the initial velocity is zero. The correct answer is option B. 2.11e+07 m/s
Step-by-step explanation:
The question entails finding the velocity of an electron after being released in an electric field for a certain duration. We start by applying Newton's second law to determine the force acting on the electron in the electric field, and then use this force to find the acceleration. Once acceleration is known, we can use kinematic equations to calculate the velocity of the electron after the given time period.
First, the force on the electron in the electric field (E) is given by F = qE, where q is the charge of the electron. Using the provided electric field (E = 337 N/C) and the fundamental charge of an electron (q = -1.602 × 10⁻¹¹ C), we can calculate the force. Since the charge of an electron is negative, the electron will accelerate in the direction opposite the electric field.
Next, we calculate the acceleration (a) of the electron using a = F/m, where m is the mass of the electron (9.109 × 10⁻³¹ kg). With the acceleration, we can now use the kinematic equation v = u + at to find the final velocity (v) of the electron, where u is the initial velocity (0 for at rest) and t is the time (57.1 ns or 57.1 × 10⁻¹⁹ s).
Therefore, the velocity of the electron after 57.1 ns is calculated using the equation v = at, as the initial velocity is zero. Substituting the calculated acceleration and time into this equation will yield the final velocity, from which we can select the appropriate answer from the given options.