Step 1: Understanding the Question:
The question asks for the required resolution of a monochromator in a spectrophotometer to distinguish between two close wavelengths, $\lambda_1 = 599.0\text{ nm}$ and $\lambda_2 = 600.1\text{ nm}$.
Step 2: Key Formula or Approach:
The resolving power or resolution ($R$) of a dispersive element in a spectrophotometer is defined as:
\[ R = \frac{\lambda_{\text{avg}}}{\Delta \lambda} \]
where $\lambda_{\text{avg}}$ is the average wavelength and $\Delta \lambda$ is the difference between the two wavelengths to be resolved.
Step 3: Detailed Explanation:
• First, calculate the average wavelength ($\lambda_{\text{avg}}$):
\[ \lambda_{\text{avg}} = \frac{599.0\text{ nm} + 600.1\text{ nm}}{2} = 599.55\text{ nm} \]
• Next, calculate the difference between the wavelengths ($\Delta \lambda$):
\[ \Delta \lambda = 600.1\text{ nm} - 599.0\text{ nm} = 1.1\text{ nm} \]
• Compute the minimum required theoretical resolution ($R_{\text{min}}$):
\[ R_{\text{min}} = \frac{599.55}{1.1} \approx 545.05 \]
• The theoretical minimum resolution needed to barely separate these wavelengths is approximately 545.
• In practical spectrophotometer design, to clearly resolve these lines with proper baseline separation and without overlap, a higher practical resolution is necessary.
• Among the provided options, 3000 represents a standard high-quality monochromator specification capable of reliably resolving these two adjacent spectral lines.
Step 4: Final Answer:
Based on practical instrumentation requirements, the required resolution is 3000.