The forbidden energy gap in Ge is 0.72,eV. Gi | vedclass

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(C) The energy of the incident photon must be at least equal to the forbidden energy gap $(E_g)$ to generate an electron-hole pair.$E_g = \frac{hc}{\l

Vedclass · Accepted Answer

$17222$

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(C) The energy of the incident photon must be at least equal to the forbidden energy gap $(E_g)$ to generate an electron-hole pair.$E_g = \frac{hc}{\lambda_{	ext{max}}}$Given $E_g = 0.72\,eV$ and $hc = 12400\,eV\cdot\mathring{A}$.Rearranging the formula for $\lambda_{	ext{max}}$:$\lambda_{	ext{max}} = \frac{hc}{E_g}$$\lambda_{	ext{max}} = \frac{12400\,eV\cdot\mathring{A}}{0.72\,eV}$$\lambda_{	ext{max}} = 17222.22\,\mathring{A}$Rounding to the nearest integer,we get $\lambda_{	ext{max}} \approx 17222\,\mathring{A}$.

The forbidden energy gap in $Ge$ is $0.72\,eV$. Given,$hc = 12400\,eV\cdot\mathring{A}$. The maximum wavelength of radiation that will generate an electron-hole pair is..........$\mathring{A}$.

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