Question

To detect light of wavelength $500nm$, the photodiode must be fabricated from, a semiconductor of minimum bandwidth of:

• A. $1.24eV$
• B. $0.62eV$
• C. $2.48eV$
• D. $3.2eV$
• E. $4.48eV$

Answer 1

Answer: (C)

$2.48eV$

Minimum bandwidth $\Delta E=h\nu =\frac{hc}{\lambda }$

where $h=$ Plank's constant , $c=$ speed of light and $\lambda =$ wavelength of light.

So, $\Delta E=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{500\times {10}^{-9}}=3.972\times {10}^{-19}J=\frac{3.972\times {10}^{-19}}{1.6\times {10}^{-19}}=2.48eV$.

where $1eV=1.6\times {10}^{-19}J$