Physics - Semiconductors Question with Solution | TestHub

Key Idea Only signals having wavelength less than threshold wavelength will be detected. Energy $$\begin{array}{rl}& \mathrm{E}=\mathrm{hv}=\mathrm{h}\frac{\mathrm{c}}{\lambda }\\ & \lambda =\frac{\mathrm{hc}}{\mathrm{E}}\end{array}$$ $$\Rightarrow \lambda =\frac{\mathrm{hc}}{\mathrm{E}}$$ Substituting the values of and in the above equation $$\lambda =\frac{6.6\times 10^{-34}\times 3\times 10^8}{2.5\times 1.6\times 10^{-19}}=5000\mathring{A}$$ As $4000\mathring{A}<5000\mathring{A}$ Signal of wavelength $4000\mathring{A}$ can be detected by the photodiode.