Answer:
Explanation:
We are given,
The band gap for a pure semiconductor is 2.1 eV. = [tex]\sf 3.365 \times 10^ {-19} \sf \ nm [/tex]
To calculate The maximum wavelength of photon which is able to create electron hole pair, we can use the formula,
[tex] \sf E_g = \dfrac{hc}{\lambda}[/tex]
where,
substitute the required values in the above formula,
[tex]\sf \lambda = \dfrac{ 6.62 \times 10^{-34} \times 3 \times 10^8 }{3.365\times 10^ {-19} } [/tex]
[tex]\sf \lambda = \dfrac{19.86 \times 10^{-26}}{3.365\times 10^ {-19} J } [/tex]
[tex] \sf \lambda = 5.90 \times 10^{-26} \times 10^{19} [/tex]
[tex] \sf \lambda = 5.90 \times 10^{7}[/tex]
[tex] \sf \lambda = 590 \ \sf nm[/tex]
Therefore, the maximum wavelength of photon which is able to create electron hole pair is nearly is 590 nm.
