Let's solve the problem step by step. ### **Step 1: Write the photoelectric equation** The Einstein photoelectric equation is: \[ hf = \phi + eV_s \] - \( h \): Planck's constant - \( f \): frequency of incident light - \( \phi \): work function of the metal - \( e \): elementary charge - \( V_s \): stopping potential The cutoff frequency \( f_c \) is when \( V_s = 0 \): \[ hf_c = \phi \] Let: - For metal A: \(\phi_A, f_{cA}, V_{sA}\) - For metal B: \(\phi_B, f_{cB}, V_{sB}\) - For metal C: \(\phi_C, f_{cC}, V_{sC}\) --- ### **Step 2: List the given relationships** 1. **Cutoff frequency of B is half of A:** \[ f_{cB} = \frac{1}{2} f_{cA} \] 2. **Work function of C is 2 eV less than A:** \[ \phi_C = \phi_A - 2~\text{eV} \] 3. **Stopping potential of B is twice that of A:** \[ V_{sB} = 2V_{sA} \] 4. **Stopping potential of C is 2.5 times that of A:** \[ V_{sC} = 2.5V_{sA} \] --- ### **Step 3: Express work functions in terms of cutoff frequency** \[ \phi_A = h f_{cA} \] \[ \phi_B = h f_{cB} = h \left(\frac{1}{2} f_{cA}\right) = \frac{1}{2} h f_{cA} = \frac{1}{2} \phi_A \] --- ### **Step 4: Write the photoelectric equation for each metal** For all metals, light with frequency \(f\) shines on them: #### **Metal A:** \[ hf = \phi_A + eV_{sA} \implies eV_{sA} = hf - \phi_A \] #### **Metal B:** \[ hf = \phi_B + eV_{sB} \implies eV_{sB} = hf - \phi_B \] #### **Metal C:** \[ hf = \phi_C + eV_{sC} \implies eV_{sC} = hf - \phi_C \] --- ### **Step 5: Use the relationships** #### **For B:** From above: \[ V_{sB} = 2V_{sA} \] \[ eV_{sB} = hf - \phi_B \] \[ eV_{sA} = hf - \phi_A \] So, \[ hf - \phi_B = 2(hf - \phi_A) \] \[ hf - \phi_B = 2hf - 2\phi_A \] \[ \phi_B = hf - 2hf + 2\phi_A = -hf + 2\phi_A \] But we also know \(\phi_B = \frac{1}{2}\phi_A\): \[ \frac{1}{2}\phi_A = -hf + 2\phi_A \] \[ -hf = \frac{1}{2}\phi_A - 2\phi_A \] \[ -hf = -\frac{3}{2}\phi_A \] \[ hf = \frac{3}{2}\phi_A \] --- ### **Step 6: Use the relationship for C** \[ V_{sC} = 2.5V_{sA} \] \[ eV_{sC} = hf - \phi_C = 2.5 eV_{sA} \] But \(eV_{sA} = hf - \phi_A\): \[ hf - \phi_C = 2.5(hf - \phi_A) \] \[ hf - \phi_C = 2.5hf - 2.5\phi_A \] \[ \phi_C = hf - 2.5hf + 2.5\phi_A \] \[ \phi_C = -1.5hf + 2.5\phi_A \] But \(\phi_C = \phi_A - 2\): \[ \phi_A - 2 = -1.5hf + 2.5\phi_A \] \[ -2 = -1.5hf + 1.5\phi_A \] \[ 1.5hf = 1.5\phi_A + 2 \] \[ hf = \phi_A + \frac{2}{1.5} \] \[ hf = \phi_A + \frac{4}{3} \] But from Step 5, \(hf = \frac{3}{2}\phi_A\): \[ \frac{3}{2}\phi_A = \phi_A + \frac{4}{3} \] \[ \frac{3}{2}\phi_A - \phi_A = \frac{4}{3} \] \[ \frac{1}{2}\phi_A = \frac{4}{3} \] \[ \phi_A = \frac{8}{3}\ \text{eV} \] --- ### **Step 7: Calculate \(hf\), then \(f\), then wavelength** From Step 5, \(hf = \frac{3}{2}\phi_A = \frac{3}{2} \times \frac{8}{3} = 4~\text{eV}\) So, energy of the photon is \(4~\text{eV}\). \[ E = hf = 4~\text{eV} = 4 \times 1.602 \times 10^{-19}~\text{J} = 6.408 \times 10^{-19}~\text{J} \] The wavelength: \[ E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E} \] Where: - \(h = 6.626 \times 10^{-34}~\text{Js}\) - \(c = 3.00 \times 10^{8}~\text{m/s}\) \[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{6.408 \times 10^{-19}} \] \[ \lambda = \frac{1.9878 \times 10^{-25}}{6.408 \times 10^{-19}} \] \[ \lambda \approx 3.10 \times 10^{-7}~\text{m} \] \[ \lambda \approx 310~\text{nm} \] --- ## **Final Answer** ### **The wavelength of the light is \(\boxed{310~\text{nm}}\).** --- #### **Summary of Steps:** 1. Used photoelectric equation and relationships provided. 2. Expressed all values in terms of work function of A. 3. Solved for \(\phi_A\) and photon energy. 4. Converted energy to wavelength. If you need a written explanation for each step, let me know!

Let's solve the problem step-by-step with detailed calculations and explanations. --- ## **Given Data and Relationships:** - The light shines on metals A, B, and C. - The cutoff frequency of B is half that of A: \[ f_{cB} = \frac{1}{2} f_{cA} \] - The work function of C is 2 eV less than that of A: \[ \phi_C = \phi_A - 2~\text{eV} \] - The stopping potential for B is twice that of A: \[ V_{sB} = 2 V_{sA} \] - The stopping potential for C is 2.5 times that of A: \[ V_{sC} = 2.5 V_{sA} \] --- ## **Step 1: Write the photoelectric equation** The Einstein photoelectric equation: \[ hf = \phi + eV_s \] - \(h\): Planck's constant - \(f\): frequency of incident light - \(\phi\): work function of the metal - \(V_s\): stopping potential - \(e\): elementary charge **At cutoff frequency**, the stopping potential \(V_s = 0\): \[ hf_c = \phi \] --- ## **Step 2: Express the work functions in terms of cutoff frequency** - For metal A: \[ \phi_A = h f_{cA} \] - For metal B: \[ \phi_B = h f_{cB} = h \times \frac{1}{2} f_{cA} = \frac{1}{2} h f_{cA} = \frac{1}{2} \phi_A \] - For metal C: \[ \phi_C = \phi_A - 2~\text{eV} \] --- ## **Step 3: Write the photoelectric equations for each metal** Assuming the incident light has frequency \(f\): ### Metal A: \[ hf = \phi_A + eV_{sA} \] (Equation 1) ### Metal B: \[ hf = \phi_B + eV_{sB} \] (Equation 2) ### Metal C: \[ hf = \phi_C + eV_{sC} \] (Equation 3) --- ## **Step 4: Use the given relationships between stopping potentials** From the problem: \[ V_{sB} = 2 V_{sA} \] \[ V_{sC} = 2.5 V_{sA} \] Express \(eV_{sB}\) and \(eV_{sC}\): \[ eV_{sB} = 2 e V_{sA} \] \[ eV_{sC} = 2.5 e V_{sA} \] --- ## **Step 5: Derive relations involving \(hf\) and work functions** ### For Metal B: From Equation 2: \[ hf = \phi_B + eV_{sB} \] Substitute \(\phi_B = \frac{1}{2} \phi_A\) and \(eV_{sB} = 2 e V_{sA}\): \[ hf = \frac{1}{2} \phi_A + 2 e V_{sA} \] (Equation 4) ### For Metal A: From Equation 1: \[ hf = \phi_A + eV_{sA} \] (Equation 5) Subtract Equation 5 from Equation 4: \[ (hf) - (hf) = \left(\frac{1}{2} \phi_A + 2 e V_{sA}\right) - (\phi_A + eV_{sA}) \] Simplify: \[ 0 = \frac{1}{2} \phi_A + 2 e V_{sA} - \phi_A - eV_{sA} \] \[ 0 = -\frac{1}{2} \phi_A + eV_{sA} \] \[ eV_{sA} = \frac{1}{2} \phi_A \] **This is a key relation:** \[ \boxed{ e V_{sA} = \frac{1}{2} \phi_A } \] Similarly, from Equation 1: \[ hf = \phi_A + eV_{sA} \] But since \(eV_{sA} = \frac{1}{2} \phi_A\), substitute: \[ hf = \phi_A + \frac{1}{2} \phi_A = \frac{3}{2} \phi_A \] --- ## **Step 6: Find the photon energy \(hf\) and work function \(\phi_A\)** \[ hf = \frac{3}{2} \phi_A \] Now, use the relation for C: \[ \phi_C = \phi_A - 2~\text{eV} \] And the photoelectric equation for C: \[ hf = \phi_C + eV_{sC} \] But \(eV_{sC} = 2.5 eV_{sA}\), and from earlier: \[ eV_{sA} = \frac{1}{2} \phi_A \] \[ \Rightarrow eV_{sC} = 2.5 \times \frac{1}{2} \phi_A = 1.25 \phi_A \] Substitute into Equation 3: \[ hf = \phi_C + eV_{sC} = (\phi_A - 2) + 1.25 \phi_A \] \[ hf = \phi_A - 2 + 1.25 \phi_A = 2.25 \phi_A - 2 \] But from earlier: \[ hf = \frac{3}{2} \phi_A \] Set equal: \[ \frac{3}{2} \phi_A = 2.25 \phi_A - 2 \] Solve for \(\phi_A\): \[ \frac{3}{2} \phi_A - 2.25 \phi_A = -2 \] \[ \left(\frac{3}{2} - 2.25\right) \phi_A = -2 \] \[ \left(1.5 - 2.25\right) \phi_A = -2 \] \[ (-0.75) \phi_A = -2 \] \[ \phi_A = \frac{-2}{-0.75} = \frac{2}{0.75} = \frac{8}{3} \approx 2.6667~\text{eV} \] --- ## **Step 7: Calculate photon energy \(hf\)** \[ hf = \frac{3}{2} \phi_A = \frac{3}{2} \times \frac{8}{3} = 4~\text{eV} \] **Photon energy:** \[ E = 4~\text{eV} \] Convert to Joules: \[ 1~\text{eV} = 1.602 \times 10^{-19}~\text{J} \] \[ E = 4 \times 1.602 \times 10^{-19} = 6.408 \times 10^{-19}~\text{J} \] --- ## **Step 8: Find the wavelength \(\lambda\)** Using: \[ E = \frac{hc}{\lambda} \] \[ \lambda = \frac{hc}{E} \] Where: - \(h = 6.626 \times 10^{-34}~\text{Js}\) - \(c = 3.00 \times 10^{8}~\text{m/s}\) Calculate: \[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{6.408 \times 10^{-19}} \] \[ \lambda = \frac{1.9878 \times 10^{-25}}{6.408 \times 10^{-19}} \] \[ \lambda \approx 3.10 \times 10^{-7}~\text{m} \] Convert to nanometers: \[ \lambda \approx 310~\text{nm} \] --- ## **Final Answer:** \[ \boxed{ \text{Wavelength of the incident light} \approx \mathbf{310~\text{nm}} } \] --- ### **Summary:** - Derived work function \(\phi_A \approx 2.67~\text{eV}\). - Photon energy \(E \approx 4~\text{eV}\). - Wavelength \(\approx 310~\text{nm}\). This completes the detailed solution.