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Here are the answers to the questions asked: Question 4 a) A photon is a quantum (or packet of energy) of electromagnetic radiation. It has no rest mass and travels at the speed of light, carrying an amount of energy proportional to its frequency. b) (i) To show the energy of each photon: The energy of a photon is given by the formula E = (hc)/(). Step 1: Identify the given values and constants. Wavelength = 620 nm = 620 × 10^-9 m. Planck's constant h = 6.63 × 10^-34 J s. Speed of light c = 3.0 × 10^8 m s^-1. Step 2: Substitute the values into the formula and calculate. E = (6.63 × 10^-34 J s) × (3.0 × 10^8 m s^-1)620 × 10^-9 m E = 1.989 × 10^-25620 × 10^-9 J E ≈ 3.208 × 10^-19 J E ≈ 3.2 × 10^-19 J The energy of each photon is indeed 3.2 × 10^-19 J. (ii) To calculate the number of photons falling on the metal surface each second: Intensity I is power per unit area, and power is energy per second. Step 1: Calculate the power of the beam. Power P = I × A. Intensity I = 5.0 × 10^2 W m^-2. Area A = 1.2 × 10^-3 m^2. P = (5.0 × 10^2 W m^-2) × (1.2 × 10^-3 m^2) P = 0.60 W Step 2: Calculate the number of photons per second. The number of photons per second N = (P)/(E). Energy per photon E = 3.2 × 10^-19 J (calculated in (i)). N = 0.60 J s^-13.2 × 10^-19 J/photon N ≈ 1.875 × 10^18 photons s^-1 The number of photons falling on the metal surface each second is approximately 1.9 × 10^18 photons s^-1. (iii) To explain if electrons will be ejected: Step 1: Convert the work function to Joules. Work function = 2.4 eV. 1 eV = 1.602 × 10^-19 J. = 2.4 × (1.602 × 10^-19 J) ≈ 3.845 × 10^-19 J Step 2: Compare the photon energy with the work function. Photon energy E = 3.2 × 10^-19 J. Work function ≈ 3.845 × 10^-19 J. Since E < (3.2 × 10^-19 J < 3.845 × 10^-19 J), the energy of the incident photons is insufficient to overcome the work function of the metal. Conclusion: No electrons will be ejected from the metal surface. Question 5 c) (i) The diagram of electric field lines and the trajectory of electrons: Electrons, being negatively charged, are deflected towards the positive plate. In the figure, they are deflected upwards, which means the upper plate is positive and the lower plate is negative. Electric field lines go from the positive plate to the negative plate (i.e., downwards). The trajectory of the electrons is parabolic. That's 2 down. 3 left today — send the next one.