Question

A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength

• A. 5500 $\mathring{A}$
• B. 6000 $\mathring{A}$
• C. 7000 $\mathring{A}$
• D. 4000 $\mathring{A}$

Answer 1

Answer: (D)

4000 $\mathring{A}$

Answer 2

Key Idea Only signals having wavelength threshold wavelength will be detected Energy $\, \, \, \, \, \, \, \, E=hv=h\frac{c}{\lambda}$ $\Rightarrow \, \, \, \, \, \, \, \, \, \, \lambda =\frac{hc}{E}$ Substituting the values of h, e and E 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.