A cyclotron produces a 100-μA beam of 15-MeV deuterons. If the cyclotron
were 100% efficient in converting electrical energy into kinetic energy of the
deuterons, what is the minimum required power input, (in kilowatts)?
To calculate the minimum required power input for the cyclotron, we need to determine the total energy converted per second (power) and then convert that into kilowatts.
Step 1: Calculate the total energy converted per second (power):
The total energy converted per second can be calculated by multiplying the beam current (I) with the beam energy (E).
Given:
- Beam current (I) = 100 μA (microamperes) = 100 x 10^-6 A
- Beam energy (E) = 15 MeV (mega-electron volts) = 15 x 10^6 eV
Energy per second (power) = Beam current (I) x Beam energy (E)
= (100 x 10^-6 A) x (15 x 10^6 eV)
To convert electron volts (eV) to joules (J), we can use the conversion factor:
1 eV = 1.6 x 10^-19 J
Therefore, the energy per second (power) in joules can be calculated as follows:
Power (in joules) = (100 x 10^-6 A) x (15 x 10^6 eV) x (1.6 x 10^-19 J/eV)
Step 2: Convert the power to kilowatts:
1 watt (W) = 1 joule (J) / 1 second (s)
1 kilowatt (kW) = 1000 watts (W)
Therefore, the power in kilowatts can be calculated as follows:
Power (in kilowatts) = Power (in joules) / 1000
Now, let's perform the calculations using the given values:
Power (in joules) = (100 x 10^-6) x (15 x 10^6) x (1.6 x 10^-19) = 2.4 x 10^-3 J
Power (in kilowatts) = (2.4 x 10^-3 J) / 1000 = 2.4 x 10^-6 kW
Therefore, the minimum required power input in kilowatts for the cyclotron is approximately 2.4 x 10^-6 kW.
To determine the minimum required power input, we need to calculate the total energy produced by the cyclotron and then convert it to power.
First, let's calculate the total energy produced by the cyclotron using the formula:
Energy = Current * Voltage * Time
Given:
Current = 100 μA = 100 * 10^(-6) A
Voltage = 15 MeV = 15 * 10^6 eV = 15 * 10^6 * 1.60218 * 10^(-19) J (1 eV = 1.60218 * 10^(-19) J)
Time = Since we are determining the power, we can assume a time of 1 second for simplicity.
Plugging in the values, we get:
Energy = (100 * 10^(-6) A) * (15 * 10^6 * 1.60218 * 10^(-19) J) * 1 s
= (100 * 15 * 1.60218) * 10^(-6) J
= (24032.7) * 10^(-6) J
= 24.0327 J
Next, we need to convert the energy to kilowatts. Since 1 kilowatt is equal to 1000 joules per second, we can use the following conversion:
Power (in kW) = Energy (in J) / 1000
Plugging in the energy value, we get:
Power = 24.0327 J / 1000
= 0.0240327 kW
Therefore, the minimum required power input for the cyclotron is approximately 0.0240327 kilowatts.