(a) Find the cutoff frequency for the photoelectric effect in copper. (b) Find the maximum energy of the ejected electrons if the copper is illuminated with light of frequency \(1.8 \times 10^{15} \mathrm{Hz}\).

Firstly, we want to find the cutoff frequency, the minimum frequency of light required to eject an electron from the surface of copper. The work function of copper, \(E\), is given as \(4.7 eV\), which we need to convert into joules (\(4.7 eV * 1.60 \times 10^{-19} J/eV = 7.52 \times 10^{-19} J\)). Then, we use the Planck-Einstein relation, \(E = hf_{cutoff}\), and we can solve for \(f_{cutoff}\) by rearranging this as \(f_{cutoff} = E/h\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34} Js\)). After substituting the necessary values, we get \(f_{cutoff} = 7.52 \times 10^{-19} J / 6.626 \times 10^{-34} Js = 1.13 \times 10^{15} Hz\).

Now, with the known cutoff frequency from step 1, we proceed to find the maximum kinetic energy of the ejected electrons when the copper is illuminated with light of frequency \(1.8 \times 10^{15} Hz\). The energy of the incident photon must be used to overcome the work function before it contributes to the kinetic energy of the photoelectron. This is modeled by \(K_{max}=E_{photon}-E_{work}\). Therefore, the incident photon energy \(E_{photon}= h*f_{incident}\) can be calculated as \(E_{photon} = 6.626 \times 10^{-34} Js * 1.8 \times 10^{15} Hz = 1.192 \times 10^{-18} J\). Following that, we calculate the maximum kinetic energy with \(K_{max} = E_{photon} - E_{work} = 1.192 \times 10^{-18} J - 7.52 \times 10^{-19} J = 4.40 \times 10^{-19} J\). To convert this to electron volts (eV), we divide by the charge of an electron \(K_{max} = 4.40 \times 10^{-19} J / 1.60 \times 10^{-19} J/eV = 2.75 eV\).