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To determine the minimum frequency of light necessary to emit electrons from titanium, we can use the Planck's equation, which states that the energy of a photon (E) is the product of Planck's constant (h) and the frequency of light (ν): \(E = hν\) We're given the minimum necessary energy to emit electrons \(E = 6.94 × 10^{-19} J\), and we know that Planck's constant \(h = 6.63 × 10^{-34} Js\). Now solve for the frequency (ν). \(ν = \frac{E}{h}\) \(ν = \frac{6.94 × 10^{-19} J}{6.63 × 10^{-34} Js} = 1.047 × 10^{15} Hz\) The minimum frequency of light necessary to emit electrons from titanium is \(1.047 × 10^{15} Hz\).

To find the wavelength of the light with the given frequency, we can use the speed of light equation: \(c = λν\) Where \(c = 3.00 × 10^8 m/s\) is the speed of light in a vacuum, λ is the wavelength, and ν is the frequency. We can solve for λ: \(λ = \frac{c}{ν}\) \(λ = \frac{3.00 × 10^8 m/s}{1.047 × 10^{15} Hz} = 2.87 × 10^{-7} m\) The wavelength of this light is \(287 nm\).

Visible light has a wavelength range of approximately \(400 nm - 700 nm\). The wavelength of the light required to emit electrons from titanium is \(287 nm\), which is shorter than the minimum wavelength of visible light (400 nm). Therefore, it is not possible to eject electrons from titanium using visible light.

We are given that the wavelength of the light irradiating the titanium is \(233 nm\). First, we need to find the energy of the photons at this wavelength. Using the speed of light equation, \(c = λν\), we can solve for the frequency (ν): \(ν = \frac{c}{λ}\) \(ν = \frac{3.00 × 10^8 m/s}{233 × 10^{-9} m} = 1.288 × 10^{15} Hz\) Now we can use the Planck's equation to find the energy of these photons: \(E_\text{photon} = hν\) \(E_\text{photon} = (6.63 × 10^{-34} Js)(1.288 × 10^{15} Hz) = 8.54 × 10^{-19} J\) Since the energy of these photons is greater than the minimum required energy to emit electrons from titanium (\(6.94 × 10^{-19} J\)), we can proceed to calculate the maximum possible kinetic energy of the emitted electrons. The excess energy from the photons will be converted into the kinetic energy of the electrons: \(K_\text{max} = E_\text{photon} - E_\text{min}\) \(K_\text{max} = 8.54 × 10^{-19} J - 6.94 × 10^{-19} J = 1.60 × 10^{-19} J\) The maximum possible kinetic energy of the emitted electrons is \(1.60 × 10^{-19} J\).