A photodiode can detect a signal if the energy of the incident photon is greater than or equal to the band gap energy of the semiconductor ($E \ge E_g$). This implies that the wavelength of the incident signal must be less than or equal to the maximum threshold wavelength ($\lambda \le \lambda_{max}$). Given, $E_g = 2.5 \text{ eV}$. We know that $E_g = \frac{hc}{\lambda_{max}}$. Using the standard approximation $hc \approx 12400 \text{ eV} \cdot \text{\AA}$ or $1242 \text{ eV} \cdot \text{nm}$: $\lambda_{max} = \frac{12400 \text{ eV} \cdot \text{\AA}}{2.5 \text{ eV}} = 4960 \text{ \AA}$ For the photodiode to detect the signal, the wavelength $\lambda$ must be $\le 4960 \text{ \AA}$. Checking the options: Option A: $6000 \text{ \AA} > 4960 \text{ \AA}$ (Cannot detect) Option B: $4000 \text{ nm} = 40000 \text{ \AA} > 4960 \text{ \AA}$ (Cannot detect) Option C: $6000 \text{ nm} = 60000 \text{ \AA} > 4960 \text{ \AA}$ (Cannot detect) Option D: $4000 \text{ \AA} < 4960 \text{ \AA}$ (Can detect) Therefore, the photodiode can detect the signal of wavelength $4000 \text{ \AA}$.