Question

Titanium metal requires a photon with a minimum energy of $6.94 \times 10^{-19} \mathrm{J}$ to emit electrons. (a) What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect? (b) What is the wavelength of this light? (c) Is it possible to eject electrons from titanium via the photoelectric effect? (b) What is the wavelength of this light? (c) Is it possible to eject electrons from titanium metal using visible light? (d) If titanium is irradiated with light of wavelength \(233 \mathrm{nm},\) what is the madimum possible kinetic energy of the emitted electrons?

Short answer

The minimum frequency of light necessary to emit electrons from titanium is \(1.047 × 10^{15} Hz\), and the corresponding wavelength is \(287 nm\). Since this is shorter than the range of visible light (\(400 nm - 700 nm\)), it's not possible to eject electrons from titanium using visible light. When titanium is irradiated with light of wavelength \(233 nm\), the maximum possible kinetic energy of the emitted electrons is \(1.60 × 10^{-19} J\).

Step-by-step solution

1. Calculate the minimum frequency of light to emit electrons from titanium

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 \(1.047 × 10^{15} Hz\).

2. Calculate the wavelength of the light

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 \(287 nm\).

3. Determine if visible light can eject electrons from titanium

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.

4. Calculate the maximum kinetic energy of emitted electrons

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 \(1.60 × 10^{-19} J\).