Introduction

This problem involves the photoelectric effect. We will use Einstein's photoelectric equation, which relates the energy of incident photons, the work function of the metal, and the maximum kinetic energy of the emitted photoelectrons. The maximum kinetic energy is related to the stopping potential.

Einstein's Photoelectric Equation

The equation is given by: $K_{max} = h\nu - \Phi$ where:

• $K_{max}$ is the maximum kinetic energy of the photoelectrons.
• $h\nu = \frac{hc}{\lambda}$ is the energy of the incident photon.
• $\Phi$ is the work function of the metal.

The maximum kinetic energy is also related to the stopping potential $V_s$ by: $K_{max} = e V_s$ where $e$ is the elementary charge.

Combining these, we get: $e V_s = \frac{hc}{\lambda} - \Phi$ Dividing the entire equation by $e$, we can express all terms in units of electron-volts (eV) for energy, or volts (V) for potential. Let $\Phi_{eV}$ be the work function in eV. The equation becomes: $V_s = \frac{hc}{e\lambda} - \frac{\Phi}{e} = \frac{hc}{e\lambda} - \Phi_{eV}$ We are given the value of the constant $\frac{hc}{e} = 1.24 \times 10^{-6} \mathrm{V} \cdot \mathrm{m}$.

Step 1: Formulate equations for the two given cases.

Case 1:

• Wavelength of incident light = $\lambda_1 = \lambda$
• Stopping potential = $V_{s1} = 6.0 \mathrm{~V}$

Substituting these values into the photoelectric equation: $6.0 = \frac{1.24 \times 10^{-6}}{\lambda} - \Phi_{eV} \quad \cdots (1)$

Case 2:

• Wavelength of incident light = $\lambda_2 = 4\lambda$
• Stopping potential = $V_{s2} = 0.6 \mathrm{~V}$
• The information about intensity being halved is irrelevant for calculating stopping potential, as stopping potential depends on the frequency (or wavelength) of incident light, not its intensity.

Substituting these values into the equation: $0.6 = \frac{1.24 \times 10^{-6}}{4\lambda} - \Phi_{eV} \quad \cdots (2)$

Step 2: Solve the system of two equations to find the wavelength $\lambda$.

We have two linear equations with two unknowns, $\lambda$ and $\Phi_{eV}$. We can eliminate $\Phi_{eV}$ by subtracting equation (2) from equation (1). $(6.0 - 0.6) = \left( \frac{1.24 \times 10^{-6}}{\lambda} - \Phi_{eV} \right) - \left( \frac{1.24 \times 10^{-6}}{4\lambda} - \Phi_{eV} \right)$ $5.4 = \frac{1.24 \times 10^{-6}}{\lambda} - \frac{1.24 \times 10^{-6}}{4\lambda}$ $5.4 = \frac{1.24 \times 10^{-6}}{\lambda} \left( 1 - \frac{1}{4} \right)$ $5.4 = \frac{1.24 \times 10^{-6}}{\lambda} \left( \frac{3}{4} \right)$ Now, we solve for $\lambda$: $\lambda = \frac{1.24 \times 10^{-6} \times 3}{5.4 \times 4}$ $\lambda = \frac{3.72 \times 10^{-6}}{21.6}$ $\lambda \approx 0.1722 \times 10^{-6} \mathrm{~m}$ $\lambda = 1.72 \times 10^{-7} \mathrm{~m}$

Step 3: Solve for the work function $\Phi_{eV}$.

We can substitute the value of $\lambda$ back into either equation (1) or (2). Let's use equation (1): $6.0 = \frac{1.24 \times 10^{-6}}{\lambda} - \Phi_{eV}$ $\Phi_{eV} = \frac{1.24 \times 10^{-6}}{1.722 \times 10^{-7}} - 6.0$ $\Phi_{eV} = \frac{1.24}{0.1722} - 6.0$ $\Phi_{eV} \approx 7.20 - 6.0$ $\Phi_{eV} = 1.20 \mathrm{~eV}$

Conclusion

The wavelength of the first source is $\lambda = 1.72 \times 10^{-7} \mathrm{~m}$ and the work function of the metal is $\Phi = 1.20 \mathrm{~eV}$.

Comparing our results with the given options:

• A: $1.72 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}$
• B: $1.72 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}$
• C: $3.78 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}$
• D: $3.78 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}$

Our calculated values match option A.