Question

An electron with energy equal to \(4.00 \cdot 10^{2} \mathrm{eV}\) and an electron with energy equal to \(2.00 \cdot 10^{2} \mathrm{eV}\) are trapped in a uniform magnetic field and move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

Short answer

Solution: As we have obtained in Step 4, the ratio of the radii of the orbits can be found by calculating the ratio of the velocities: Ratio: \(\frac{r_1}{r_2} = \frac{v_1}{v_2}\) Using the velocities we calculated in Step 2: \(v_1 \approx \sqrt{\frac{2 \cdot (4.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\, J)}{9.11 \cdot 10^{-31}\, kg}}\) \(v_2 \approx \sqrt{\frac{2 \cdot (2.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\, J)}{9.11 \cdot 10^{-31}\, kg}}\) Now substituting the values into the ratio formula: Ratio: \(\frac{r_1}{r_2} = \frac{\sqrt{\frac{2 \cdot (4.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\, J)}{9.11 \cdot 10^{-31}\, kg}}}{\sqrt{\frac{2 \cdot (2.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\, J)}{9.11 \cdot 10^{-31}\, kg}}}\) The square root and the denominator can be simplified: Ratio: \(\frac{r_1}{r_2} = \frac{\sqrt{4.00 \cdot 10^2}}{\sqrt{2.00 \cdot 10^2}} = \frac{20}{10}\) Therefore, the ratio of the radii of the orbits is 2:1.

Step-by-step solution

Convert energy to joules

First, we need to convert the given energies from electron-volts to joules. The conversion factor is \(1\,\mathrm{eV} = 1.602 \cdot 10^{-19} \mathrm{J}\). Energy of the first electron: \(E_1 = 4.00 \cdot 10^2 \,\mathrm{eV} = 4.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\,\mathrm{J}\) Energy of the second electron: \(E_2 = 2.00 \cdot 10^2\, \mathrm{eV} = 2.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19} \, \mathrm{J}\)

Calculate velocities

Next, using the kinetic energy formula, we will find the velocities of the electrons. \(E = \frac{1}{2}mv^2\) Solve for \(v\): \(v = \sqrt{\frac{2E}{m}}\) Where \(m = 9.11 \cdot 10^{-31} \, \mathrm{kg}\) is the mass of an electron. Velocity of the first electron: \(v_1 = \sqrt{\frac{2 \cdot (4.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\, \mathrm{J})}{9.11 \cdot 10^{-31} \,\mathrm{kg}}}\) Velocity of the second electron: \(v_2 = \sqrt{\frac{2 \cdot (2.00 \cdot 10^2 \cdot 1.602 \cdot 10^{-19}\, \mathrm{J})}{9.11 \cdot 10^{-31}\, \mathrm{kg}}}\)

Calculate radii of the orbits

Now, we will use the following relation between velocity, charge, magnetic field, and radius of the orbit: \(v = \frac{qB}{mv} \cdot r\) Solve for \(r\): \(r = \frac{mv}{qB}\) Where \(q = 1.602 \cdot 10^{-19}\, \mathrm{C}\) is the charge of an electron and \(B\) is the magnetic field. Radius of the first electron's orbit: \(r_1 = \frac{(9.11 \cdot 10^{-31}\, \mathrm{kg}) \cdot v_1 }{(1.602 \cdot 10^{-19}\, \mathrm{C}) \cdot B}\) Radius of the second electron's orbit: \(r_2 = \frac{(9.11 \cdot 10^{-31}\, \mathrm{kg}) \cdot v_2 }{(1.602 \cdot 10^{-19}\, \mathrm{C}) \cdot B}\)

Calculate the ratio of the radii

Finally, we will find the ratio of the orbits' radii: Ratio: \(\frac{r_1}{r_2} = \frac{\frac{(9.11 \cdot 10^{-31}\, \mathrm{kg}) \cdot v_1 }{(1.602 \cdot 10^{-19}\, \mathrm{C}) \cdot B}}{\frac{(9.11 \cdot 10^{-31}\, \mathrm{kg}) \cdot v_2 }{(1.602 \cdot 10^{-19}\, \mathrm{C}) \cdot B}}\) Simplifying the expression, we get: Ratio: \(\frac{r_1}{r_2} = \frac{v_1}{v_2}\) Now, we can calculate the values of \(v_1\) and \(v_2\) from "Step 2" and find the ratio of their orbits' radii.