Step 1: Recall the photoelectric equation.
When light hits a metal, the freed electron's largest kinetic energy is the photon energy minus the work needed to escape: \[ K_{\max} = h\nu - h\nu_0, \] where $\nu_0$ is the threshold frequency.
Step 2: First case with frequency $2\nu_0$.
The kinetic energy is \[ K_1 = h(2\nu_0) - h\nu_0 = h\nu_0. \] Since $K = \tfrac{1}{2}mv^2$, we get $\tfrac{1}{2}mv_1^2 = h\nu_0$.
Step 3: Second case with frequency $5\nu_0$.
Now \[ K_2 = h(5\nu_0) - h\nu_0 = 4h\nu_0, \] so $\tfrac{1}{2}mv_2^2 = 4h\nu_0$.
Step 4: Divide the two energies.
Dividing cancels the constants: \[ \frac{v_1^2}{v_2^2} = \frac{h\nu_0}{4h\nu_0} = \frac{1}{4}. \]
Step 5: Take the square root.
\[ \frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}. \]
Step 6: State the answer.
So the speeds are in the ratio one to two. \[ \boxed{1:2} \]