The cyclotron could produce beams of alpha particles with kinetic energy up to 16 MeV.Calculate the magnetic flux density required by the cyclotron when alpha particleswith kinetic energy of 16 MeV are produced.diameter of cyclotron = 0.94 mmass of alpha particle = 6.6 × 10−27 k
Question
The cyclotron could produce beams of alpha particles with kinetic energy up to 16 MeV.Calculate the magnetic flux density required by the cyclotron when alpha particleswith kinetic energy of 16 MeV are produced.diameter of cyclotron = 0.94 mmass of alpha particle = 6.6 × 10−27 k
Solution
To solve this problem, we need to use the formula for the kinetic energy of a particle in a cyclotron, which is given by:
K.E = qVB
where: K.E is the kinetic energy, q is the charge of the particle, V is the voltage, and B is the magnetic flux density.
We are given that the kinetic energy is 16 MeV (Mega electron Volts), which we need to convert to Joules. We know that 1 eV = 1.6 x 10^-19 Joules, so 16 MeV = 16 x 10^6 x 1.6 x 10^-19 Joules = 2.56 x 10^-13 Joules.
We also know that an alpha particle has a charge of 2e, where e is the elementary charge, 1.6 x 10^-19 Coulombs. So q = 2 x 1.6 x 10^-19 = 3.2 x 10^-19 Coulombs.
We can rearrange the formula for kinetic energy to solve for B:
B = K.E / (qV)
We are not given the voltage V, but we can assume that the alpha particle makes one full revolution in the cyclotron, so the voltage is the kinetic energy divided by the charge, V = K.E / q = 2.56 x 10^-13 / 3.2 x 10^-19 = 800,000 Volts.
Substituting these values into the formula gives:
B = 2.56 x 10^-13 / (3.2 x 10^-19 x 800,000) = 1 Tesla.
So the magnetic flux density required by the cyclotron when alpha particles with kinetic energy of 16 MeV are produced is 1 Tesla.
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