Question

A p-n photodiode is manufactured from a semiconductor with gap of 3.1 eV. Which of the following wavelengths can be detected by it?

• A. 4000 $\stackrel{\circ }{A}$
• B. 3900 $\stackrel{\circ }{A}$
• C. 4200 $\stackrel{\circ }{A}$
• D. Both (1) & (2)

Answer 1

The correct option is A 4000 $\stackrel{\circ }{A}$

For a $P-N$ Junction Diode :

$E\left(ev\right)=h\nu$

$=\frac{hc}{\lambda }$

$=\frac{12400eV-A^o}{\lambda \left(A^o\right)}$

Since given $E\left(ev\right)=3.1eV$

$\Rightarrow 3.1=\frac{12400}{\lambda }$

$\Rightarrow \lambda =\frac{12400}{3.1}$

$=4000A^o$