Final answer:
The speed of an electron after being accelerated from rest through a potential difference of 350 V is approximately 3.31 × 10^7 m/s, which is about 11.03% of the speed of light.
Step-by-step explanation:
To calculate the speed of an electron after being accelerated from rest through a potential difference of 350 V, we can use the principle of energy conservation. The kinetic energy (KE) gained by the electron is equal to the work done on it by the electric field, which is equal to the charge of the electron (e) multiplied by the potential difference (Δν).
The kinetic energy gained by the electron can be expressed as KE = ½ mv², where m is the mass of the electron (9.11 × 10-31 kg) and v is the velocity of the electron. Equating the kinetic energy to the work done gives us:
½ mv² = eΔν
Plugging in the known values (e = 1.60 × 10-19 C and Δν = 350 V) and solving for v gives us:
v = sqrt((2 * e * Δν) / m)
After calculating, we find that v ≈ 3.31 × 107 m/s. To express this as a percentage of the speed of light (c = 3.0 × 108 m/s), we calculate (v/c) × 100%.
Therefore, the percentage of the speed of light is approximately 11.03%.