Photodiode

A photodiode can detect light only if the photon energy is at least equal to the band gap: $h\nu \geq E_g$, i.e., $\lambda \leq \frac{hc}{E_g}$
Using hc = 12400 eV Å: $\lambda_{max} = \frac{12400}{2.5} = 4960$ Å
So only wavelengths up to 4960 Å (496 nm) can be detected.
6000 Å, 4000 nm (40000 Å) and 6000 nm (60000 Å) are all longer than 4960 Å; their photons have less energy than 2.5 eV and cannot be detected.
Hence it can detect a signal of wavelength 4960 Å.
Note: the source prints this option as '496 Å'; the standard value in the original paper, 4960 Å, is used here.

A photon is absorbed only if its energy is at least equal to the band gap: $h\nu \geq E_g$.
The minimum frequency corresponds to $E = h\nu = 2$ eV.
$2 \times 1.6 \times 10^{-19} = 6.6 \times 10^{-34} \times \nu$
$\nu = \dfrac{3.2 \times 10^{-19}}{6.6 \times 10^{-34}}$
$\nu = 4.85 \times 10^{14}$ Hz $\approx 5 \times 10^{14}$ Hz
So the minimum frequency that can be absorbed is nearly $5 \times 10^{14}$ Hz.