A proton, a deutron and an a-particle having equal kinetic energies. Compare the radius of their paths when a normal magnetic field is applied
e force due to the magnetic field is ALWAYS perpendicular to the velocity, which means it is a centripetal force...we know the centripetal force is given by:
F = mv²/r
-->
Furthermore, we know the force of the magnetic field is:
F = qvB
...assuming v and B are perpendicular...the resulting force will be in a direction perpendicular to BOTH the magnetic field AND the velocity. This is what gives the circular motion in 2 dimensions...
So you have:
qvB = mv²/r
--> solve for r
r = mv² / (qvB) = mv/(qB)
You can find v from the initial kinetic energy:
T = ½mv² --> v = √(2T/m)
-->
r = mv/(qB) = √(2Tm) / (qB)
So the proton and deuteron have the same charge: +1. The alpha particle has charge +2. Assuming the neutron and proton weight the same (the neutron is slightly more massive in reality), then we have:
m_proton = 1, q_proton = 1
m_deuteron = 2, q_deuteron = 1
m_alpha = 4, q_alpha = 2
r_proton = √(2Tm) / (qB) = √2 * √T / B
r_deuteron = √2 * √(T * 2) = 2√T /B
r_alpha = √2 * √(T*4) / 2 = 2√2 * √T / 2 = √2 * √T / B
So the proton and alpha particle have the same radius (this is because the alpha particle has twice the charge but 2² = 4 times the mass), while the deuteron's radius is √2 times larger than the proton and alpha particle.
F = mv²/r
-->
Furthermore, we know the force of the magnetic field is:
F = qvB
...assuming v and B are perpendicular...the resulting force will be in a direction perpendicular to BOTH the magnetic field AND the velocity. This is what gives the circular motion in 2 dimensions...
So you have:
qvB = mv²/r
--> solve for r
r = mv² / (qvB) = mv/(qB)
You can find v from the initial kinetic energy:
T = ½mv² --> v = √(2T/m)
-->
r = mv/(qB) = √(2Tm) / (qB)
So the proton and deuteron have the same charge: +1. The alpha particle has charge +2. Assuming the neutron and proton weight the same (the neutron is slightly more massive in reality), then we have:
m_proton = 1, q_proton = 1
m_deuteron = 2, q_deuteron = 1
m_alpha = 4, q_alpha = 2
r_proton = √(2Tm) / (qB) = √2 * √T / B
r_deuteron = √2 * √(T * 2) = 2√T /B
r_alpha = √2 * √(T*4) / 2 = 2√2 * √T / 2 = √2 * √T / B
So the proton and alpha particle have the same radius (this is because the alpha particle has twice the charge but 2² = 4 times the mass), while the deuteron's radius is √2 times larger than the proton and alpha particle.