Question Detail
Question
Calculate the velocity of the electron emitted from the cathode to the anode of a cathode ray tube when a voltage of 50 kV is applied.
Correct Answer
Option C is the correct answer.
Detailed Explanation
The kinetic energy gained by the electron is eV = ½mv². V = 50 kV = 50,000 V. e = 1.602 × 10^-19 C, m = 9.11 × 10^-31 kg. v = √(2eV/m) = √(2 × 1.602e-19 × 50000 / 9.11e-31) = √(1.602e-14 / 9.11e-31) = √(1.758 × 10^16) = 1.326 × 10^8 m/s ≈ 1.3 × 10^8 m/s.
Hint
Use the energy conservation: eV = ½mv².