A p-n photodiode is fabricated from a semiconductor with a band gap of $2.5\text{eV}$. It can detect a signal of wavelength:

For a photodiode to detect a signal, the energy of the incident photon ($E$) must be greater than or equal to the band gap energy ($E_g$).
$$E\ge E_g$$
$$\frac{hc}{\lambda }\ge E_g⟹\lambda \le \frac{hc}{E_g}$$
Using the approximation $hc\approx 12400\text{eV A˚}$:
$$\lambda \le \frac{12400}{2.5}=4960\text{A˚}$$
Let's check the options:$4000\text{nm}=40000\text{A˚}$ (Too large)$6000\text{nm}=60000\text{A˚}$ (Too large)$4000\text{A˚}<4960\text{A˚}$ (Detectable)$6000\text{A˚}>4960\text{A˚}$ (Not detectable)Only $4000\text{A˚}$ is smaller than the threshold wavelength.