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Q.

The forbidden energy gap in Ge is 0.72 eV. Given hc=12400eV-A. the maximum wavelength of radiation that will generate electron hole pair is

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a

17222 $\overset{0}{A}$

b

1722 $\overset{0}{A}$

c

172220 $\overset{0}{A}$

d

172.2 $\overset{0}{A}$

answer is C.

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Detailed Solution

Energy gap ,

$\begin{array}{l} E_{g} = \frac{h c}{λ} \\ λ = \frac{h c}{E_{g}} = \frac{12400 e V \overset{0}{A}}{0.72 e V} = 17222 \overset{0}{A} \end{array}$

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