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.
a) To calculate the kinetic energy of the proton at point A, we can use the formula:
Kinetic energy (KE) = (1/2) * mass * velocity^2
First, let's convert the speed from kilometers per second (km/s) to meters per second (m/s):
Speed at A = 2000 km/s
Speed at A = 2000 km/s * 1000 m/km
Speed at A = 2,000,000 m/s
The mass of a proton is approximately 1.67 x 10^-27 kg.
Now we can calculate the kinetic energy:
KE = (1/2) * (1.67 x 10^-27 kg) * (2,000,000 m/s)^2
KE = 1/2 * 1.67 x 10^-27 * 4 x 10^12
KE ≈ 3.34 x 10^-15 joules
To convert this value to electron volts (eV), we can use the conversion factor: 1 eV = 1.60218 x 10^-19 J. Therefore:
KE (eV) = (3.34 x 10^-15 J) / (1.60218 x 10^-19 J/eV)
KE (eV) ≈ 20855.38 eV
So, at point A, the kinetic energy of the proton is approximately 3.34 x 10^-15 joules or 20855.38 electron volts (eV).
b) To calculate the voltage (U) required between points A and B, we need to consider the change in kinetic energy and the charge of the proton.
The change in kinetic energy is given by the formula:
Change in KE = KE(B) - KE(A)
Since we want the proton to pass point B with a speed of 1000 km/s, we calculate the kinetic energy at point B using the same formula as in part a, with the new speed:
Speed at B = 10000 km/s
Speed at B = 10000 km/s * 1000 m/km
Speed at B = 10,000,000 m/s
KE(B) = (1/2) * (1.67 x 10^-27 kg) * (10,000,000 m/s)^2
KE(B) ≈ 8.35 x 10^-13 joules
The change in kinetic energy is then:
Change in KE = (8.35 x 10^-13 J) - (3.34 x 10^-15 J)
Change in KE ≈ 8.32 x 10^-13 joules
Since the charge of a proton is 1.60218 x 10^-19 coulombs, we can calculate the voltage using:
Change in KE = charge * voltage
Therefore:
8.32 x 10^-13 J = (1.60218 x 10^-19 C) * U
Solving for U:
U ≈ (8.32 x 10^-13 J) / (1.60218 x 10^-19 C)
U ≈ 5197.38 volts
So, to make the proton pass point B at a speed of 10000 km/s, a positive voltage of approximately 5197.38 volts should be applied between points A and B. The positive voltage means that the positive terminal should be at point B and the negative terminal at point A.