Question

 Singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kV and enter a uniform magnetic field of 1.20 T directed perpendicular to their velocities. (a) Determine the radius of their circular path. (b) Repeat for uranium-235 ions. What If? How does the ratio of these path radii depend on the accelerating voltage? On the magnitude of the magnetic field? 

Answer 1

a) Using the Newton’s Second Law of Motion we can write:

$$qvB=\dfrac{mv^2}{r},$$ $$r=\dfrac{mv}{qB}.$$

We can find the velocity of the ions from the law of conservation of energy:

$$qV=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2qV}{m}}.$$

Substituting $v$ into the formula for the radius of the uranium-238/235 ions orbit, we get:

$$r=\dfrac{\sqrt{2Vm}}{\sqrt{q}B}.$$

Then, we can find the radius of uranium-238 ions circular path:

$$r_{238}=\dfrac{\sqrt{2\cdot2\cdot10^3\ V\cdot238.05\cdot1.66\cdot10^{-27}\ kg}}{\sqrt{1.602\cdot10^{-19}\ C}\cdot1.20\ T}=0.08278\ m.$$

b) Let's repeat the calculations for uranium-235 ions:

$$r_{235}=\dfrac{\sqrt{2\cdot2\cdot10^3\ V\cdot235.04\cdot1.66\cdot10^{-27}\ kg}}{\sqrt{1.602\cdot10^{-19}\ C}\cdot1.20\ T}=0.08225\ m.$$

c) Let's find the ratio of the path of uranium-235/238 radii:

$$\dfrac{r_1}{r_2}=\dfrac{\dfrac{\sqrt{2Vm_1}}{\sqrt{q}B}}{\dfrac{\sqrt{2Vm_2}}{\sqrt{q}B}}=\sqrt{\dfrac{m_1}{m_2}}.$$

As we can see from the formula, the ratio of these path radii doesn't depend on the accelerating voltage. Also, it doesn't depend on the magnitude of the magnetic field.