The forbidden energy gap of Ge is 0.75 eV. T | vedclass

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(B) The energy of the incident photon must be at least equal to the forbidden energy gap $(E_g)$ to create an electron-hole pair.$E = E_g = 0.75 \ eV$

Vedclass · Accepted Answer

$16500$

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(B) The energy of the incident photon must be at least equal to the forbidden energy gap $(E_g)$ to create an electron-hole pair.$E = E_g = 0.75 \ eV$Using the relation between energy $(E)$ in $eV$ and wavelength $(\lambda)$ in $\mathring{A}$:$E = \frac{12400}{\lambda (\mathring{A})}$Rearranging for $\lambda$:$\lambda = \frac{12400}{E_g} = \frac{12400}{0.75} \ \mathring{A}$$\lambda = 16533.33 \ \mathring{A} \approx 16500 \ \mathring{A}$Thus,the maximum wavelength is $16500 \ \mathring{A}$.

The forbidden energy gap of $Ge$ is $0.75 \ eV$. The maximum wavelength of incident photon radiation that can generate an electron-hole pair in a $Ge$ semiconductor is ........... $\mathring{A}$.

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