Titanium metal requires light with a maximum wavelength of 286 nm to emit electrons. (a) What is the minimum energy of the photons necessary to emit electrons from titanium via the photoelectric effect? (b) If titanium is irradiated with light of wavelength 276 nm, what is the maximum possible kinetic energy of the emitted electrons?

Verified step by step guidance

1. Step 1: Understand the photoelectric effect, which states that electrons are emitted from a material when it is exposed to light of a certain minimum frequency or maximum wavelength. The energy of the photons must be greater than or equal to the work function of the material.
2. Step 2: Use the equation for the energy of a photon, E = \frac{hc}{\lambda}, where h is Planck's constant (6.626 \times 10^{-34} \text{ J s}), c is the speed of light (3.00 \times 10^8 \text{ m/s}), and \lambda is the wavelength of the light. Calculate the minimum energy of the photons using the maximum wavelength of 286 nm.
3. Step 3: Convert the wavelength from nanometers to meters for calculation purposes. Remember that 1 nm = 1 \times 10^{-9} m.
4. Step 4: For part (b), calculate the energy of the photons with a wavelength of 276 nm using the same formula, E = \frac{hc}{\lambda}.
5. Step 5: Determine the maximum kinetic energy of the emitted electrons by using the equation KE_{max} = E_{photon} - \phi, where \phi is the work function of titanium, which is the energy calculated in Step 2.