Solution

When an electron moves in a magnetic field, it experiences the Lorentz force, which causes it to move in a circular trajectory. The Lorentz force for an electron moving with velocity $v$ perpendicular to the magnetic field $B$ is:
$$F = evB$$

The centripetal acceleration for circular motion is described by the expression:
$$a = \frac{v^2}{R}$$

Let us write Newton's second law:
$$evB = \frac{m_e v^2}{R}$$

From this equation, we can express the trajectory radius $R$:
$$R = \frac{m_e v}{eB} \quad (1)$$

The kinetic energy of the first electron:
$$K_1 = \frac{1}{2}m_e v^2$$

From here, we express the velocity:
$$v = \sqrt{\frac{2K_1}{m_e}}$$

Substitute into (1):
$$R_1 = \frac{m_e \sqrt{\frac{2K_1}{m_e}}}{eB} = \frac{\sqrt{2K_1 m_e}}{eB} \quad (2)$$

Similarly, for the electron with kinetic energy $K_2$:
$$R_2 = \frac{\sqrt{2K_2 m_e}}{eB} \quad (3)$$

Let us take the ratio of the radii:
$$\frac{R_1}{R_2} = \sqrt{\frac{K_1}{K_2}}$$