Final answer:
The threshold frequency of the metal is determined by the condition when the kinetic energy of ejected electrons is approximately zero.
Step-by-step explanation:
To determine the threshold frequency of the metal we can use the photoelectric effect which states that the kinetic energy of ejected electrons is given by the energy of the incoming photons minus the work function (or threshold energy) of the metal. The equation for this relationship is KkE = hf - Φ, where KkE is the kinetic energy of the electron, ˆf is the energy of the photon (h is Planck's constant and f is the frequency of the light), and Φ is the work function of the metal.
In the question, you are given that no electrons are emitted with λ2, which corresponds to the condition when the photon energy is equal to the work function of the metal (Φ). The energy of the photons can be calculated using the equation E = hf, where h is Planck's constant (6.626×10-34 Js) and f is the frequency of the light. To find the threshold frequency, you need to use the wavelength at which the kinetic energy of the ejected electrons is approximately zero. This will give you the frequency at which the incoming photon energy is just enough to eject an electron but doesn't provide any excess kinetic energy.
Unfortunately, you have not provided the corresponding wavelengths for each condition, λ1, λ2, and λ3, thus I cannot calculate the exact threshold frequency for you. I would need the wavelength associated with λ2 to complete this task.