A photodiode fabricated from lead Selenide $[\mathrm{PbSe}]$ has a band gap energy of 0.27 eV . What is the wavelength of the signal the photodiode can detect?

The energy of a photon is given by the relation $E = \dfrac{hc}{\lambda}$, where $E$ is the band gap energy, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

Given $E = 0.27 \text{ eV}$.

Using the relation $\lambda = \dfrac{hc}{E}$, we substitute the values $hc \approx 1240 \text{ eV} \cdot \text{nm}$.

$\lambda = \dfrac{1240 \text{ eV} \cdot \text{nm}}{0.27 \text{ eV}}$

$\lambda \approx 4592.59 \text{ nm}$.

Using a more precise value for $hc = 1242 \text{ eV} \cdot \text{nm}$:

$\lambda = \dfrac{1242}{0.27} = 4600 \text{ nm}$.

Comparing with the given options, the closest value is $4603 \text{ nm}$.

Answer: 4603 nm