P-type semiconductor has acceptor levels 57 , | vedclass

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Question

(C) The energy required to create a hole is equal to the energy gap between the acceptor level and the valence band,which is $E = 57 \, meV = 57 	ime

Vedclass · Accepted Answer

$217100 \, \mathring{A}$

Vedclass · Answer

(C) The energy required to create a hole is equal to the energy gap between the acceptor level and the valence band,which is $E = 57 \, meV = 57 	imes 10^{-3} \, eV$.Using the relation between energy in $eV$ and wavelength in $\mathring{A}$:$E = \frac{12400}{\lambda (\mathring{A})}$Rearranging for wavelength:$\lambda = \frac{12400}{E(eV)} = \frac{12400}{57 	imes 10^{-3}} \, \mathring{A}$$\lambda \approx 217543 \, \mathring{A}$.Rounding to the nearest provided option,the correct value is $217100 \, \mathring{A}$.

$P$-type semiconductor has acceptor levels $57 \, meV$ above the valence band. The maximum wavelength of light required to create a hole is (Planck's constant $h = 6.6 \times 10^{-34} \, J-s$)

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