### Photoelectric Effect Problem**Problem Statement:**Electrons are ejected from a metallic surface with speeds ranging up to \( 4.60 \times 10^5 \, \text{m/s} \) when light with a wavelength of 625 nm is used.(a) What is the work function of the surface?(b) What is the cutoff frequency for this surface?**Solution Outline:**To solve for the work function and the cutoff frequency, we will use principles from the photoelectric effect and related equations derived from quantum mechanics. - **Step 1: Determine the energy of the incident photons.**- **Step 2: Use the maximum kinetic energy of the ejected electrons to find the work function.**- **Step 3: Calculate the cutoff frequency using the work function.**### Graphs and DiagramsThis problem does not contain any graphs or diagrams. ### Detailed Solutions:#### (a) Determining the Work Function1. **Energy of Incident Photons:**\[ E = \frac{hc}{\lambda} \] Where: - \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \) - \( c \) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \) - \( \lambda \) is the wavelength of the incident light \( (625 \, \text{nm} = 625 \times 10^{-9} \, \text{m}) \) \[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{625 \times 10^{-9}} \]\[ E = 3.18 \times 10^{-19} \, \text{J} \]2. **Maximum Kinetic Energy of the Ejected Electrons:**\[ KE = \frac{1}{2}mv^2 \] Where: - \( m \) is the electron mass \( (9.11 \times 10^{-31} \, \text{kg}) \) - \( v \) is the maximum speed of the ejected electrons \( (4.60 \times 10^5 \, \text{m/s}) \) \[ KE = \

When some materials are exposed to electromagnetic radiation, they release electrons. This process…

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Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?