Final answer:
No, photoelectrons will not be produced when visible light shines on the surface of potassium. The velocity of the ejected electrons cannot be determined without additional information, such as the angle at which the electrons are ejected.
Step-by-step explanation:
When visible light shines on the surface of potassium, photoelectrons will be produced if the energy of the light exceeds the minimum energy required for the photoelectric effect. The minimum energy required to cause the photoelectric effect in potassium is 2.0 eV.
The velocity of the ejected electrons can be calculated using the equation KE = 1/2mv^2, where KE is the kinetic energy of the electrons and m is the mass of the electrons. To find the velocity, we need to know the mass of the electrons which is approximately 9.11x10^-31 kg.
Let's assume that the photons have a wavelength of 520 nm, corresponding to green light. Using the energy equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.63x10^-34 J·s), c is the speed of light (3x10^8 m/s), and λ is the wavelength of the photons:
E = (6.63x10^-34 J·s)(3x10^8 m/s)/(520x10^-9 m) = 3.81x10^-19 J
Since the minimum energy required for the photoelectric effect in potassium is 2.0 eV, and 1 eV is approximately equal to 1.6x10^-19 J, the energy of the photons is not enough to cause the photoelectric effect. Therefore, no photoelectrons will be produced.
It's important to note that the equation KE = 1/2mv^2 gives the maximum velocity of the ejected electrons. To calculate the actual velocity, more information is needed, such as the angle at which the electrons are ejected.