Question

Energy gap in pn photodiode is 2.8 ev. Can it detect a wavelength of 6000 nm

Answer 1

Answer:

No, it cannot detect a wavelength of 6000 nm.

Explanation:

Eg= 2.8 eV

λ(wavelength)= 6000 nm

E= hc/λ

E=(6.63x10^-34X3X10^8)/(6000*10^-9)

E=0.207 eV

Eg>E

Therefore, it cannot detect a wavelength of 6000 nm.

Answer 2

It cannot detect the given wavelength.

Explanation:

We are given that,

Energy gap in pn photo diode, $E_g=2.8\ eV$

We need to say that the phodiode can detect or not a wavelength of 6000 nm. For this first find the energy of he photodiode at this wavelength as :

$E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6000\times 10^{-9}}\\\\E=3.31\times 10^{-20}\ J$

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

$E=\dfrac{3.31\times 10^{-20}}{1.6\times 10^{-19}}\\\\E=0.206\ eV$

The energy of a photo diode is 0.206 eV. But the energy gap in pn photo diode is 2.8 eV. Hence, it cannot detect the given wavelength.

Learn more,

Energy gap in pn photodiode is 2.8 ev. Can it detect a wavelength of 6000 nm ?

https://brainly.in/question/6096117