Problem 21.10
A proton with an initial speed of 800000 m/s2 is brought to rest by an electric field.
Part B -
What was the potential difference that stopped the proton?
Part C -
What was the initial kinetic energy of the proton, in electron volts?
To find the potential difference that stopped the proton in Part B, we can use the formula:
Potential difference = (Final kinetic energy - Initial kinetic energy) / Charge of the proton
The charge of a proton is +1.6 x 10^-19 coulombs.
Since the proton is brought to rest, its final kinetic energy is zero. Therefore:
Potential difference = (0 - Initial kinetic energy) / Charge of the proton
To find the initial kinetic energy of the proton in electron volts in Part C, we can use the formula:
Initial kinetic energy (in electron volts) = Initial kinetic energy (in joules) / (1.6 x 10^-19 joules/electron volt)
Let's calculate each part step by step:
Step 1: Finding the potential difference in Part B
Potential difference = (0 - Initial kinetic energy) / Charge of the proton
Potential difference = - Initial kinetic energy / Charge of the proton
Step 2: Finding the initial kinetic energy in Part C
Initial kinetic energy (in electron volts) = Initial kinetic energy (in joules) / (1.6 x 10^-19 joules/electron volt)
To calculate the potential difference and initial kinetic energy, we need to know the mass of the proton. Could you provide the mass of the proton?