A proton is moving in vacuum along a straight path from A to B .It passes point A at the speed of 2000km/s.
a)Calculate in J and in eV the kinetic energy of the proton at A.
b)Calculate the value of the positive voltage U that should be applied between A and B so that the proton passes B at the speed of 10000km/s. Specify how this voltage should be applied.
To calculate the kinetic energy of the proton at point A, we can use the formula:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
a) To calculate the kinetic energy in joules (J), we need to know the mass of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg.
Given:
Mass of proton (m) = 1.67 x 10^-27 kg
Velocity at A (vA) = 2000 km/s = 2 x 10^6 m/s
Using the formula for kinetic energy, we get:
KE = (1/2) * m * vA^2
= (1/2) * (1.67 x 10^-27) * (2 x 10^6)^2
≈ 3.34 x 10^-19 J
So, the kinetic energy of the proton at point A is approximately 3.34 x 10^-19 joules (J).
Now, to calculate the kinetic energy in electron volts (eV), we'll use the conversion factor:
1 electron volt = 1.6 x 10^-19 J
Using this conversion factor, we can convert the kinetic energy from joules to electron volts:
KE (in eV) = KE (in J) / (1.6 x 10^-19 J)
= (3.34 x 10^-19) / (1.6 x 10^-19)
≈ 2.09 eV
Therefore, the kinetic energy of the proton at point A is approximately 2.09 electron volts (eV).
b) To calculate the voltage (U) that should be applied between A and B so that the proton passes B at a speed of 10000 km/s = 1 x 10^7 m/s, we can equate the initial kinetic energy at A with the final kinetic energy at B and solve for U.
Using the formula for kinetic energy:
KE = (1/2) * mass * velocity^2
Initial kinetic energy at A = (1/2) * mass * vA^2
Final kinetic energy at B = (1/2) * mass * vB^2
Given:
Mass of proton (m) = 1.67 x 10^-27 kg
Velocity at A (vA) = 2 x 10^6 m/s
Velocity at B (vB) = 1 x 10^7 m/s
Equating the initial and final kinetic energies, we get:
(1/2) * m * vA^2 = (1/2) * m * vB^2
Simplifying and solving for vB^2, we get:
vB^2 = (vA^2 * m) / m
vB^2 = vA^2
Substituting the values, we have:
(1 x 10^7)^2 = (2 x 10^6)^2
Simplifying, we find:
1 x 10^14 = 4 x 10^12
Now, to calculate the voltage (U), we'll use the formula:
Voltage (U) = KE / (charge of proton)
Since the proton has a charge of +1.6 x 10^-19 C, substituting the values, we get:
U = KE (in J) / (1.6 x 10^-19 C)
= 3.34 x 10^-19 / (1.6 x 10^-19)
≈ 2.09 V
Therefore, the voltage (U) that should be applied between A and B is approximately 2.09 volts (V). The positive voltage (U) should be applied in such a way that point B is at a higher potential (positive) than point A.