When 550-nm light passes through a thin slit and then travels to a screen, the first-order dark fringe in the interference pattern is at \(32.5^{\circ}\) from the center of the screen. When a beam of electrons, each having kinetic energy equal to the energy of the photons in the \(550-\mathrm{nm}\) beam, passes through the slit and travels to the screen, what is the smallest angle from the screen center at which no electrons are found?