A 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, $E = 57 \; meV = 57 	imes 10^{-3

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, $E = 57 \; meV = 57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} \; J$.Using the relation $E = \frac{hc}{\lambda}$, the maximum wavelength $\lambda$ is given by $\lambda = \frac{hc}{E}$.Substituting the values: $\lambda = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19}} \; m$.$\lambda = \frac{19.8 	imes 10^{-26}}{91.2 	imes 10^{-22}} \; m \approx 0.2171 	imes 10^{-4} \; m$.Converting to $\mathring{A}$s $(\mathring{A})$: $\lambda = 0.2171 	imes 10^{-4} 	imes 10^{10} \; \mathring{A} = 217100 \; \mathring{A}$.

$A$ $P$-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 $h = 6.6 \times 10^{-34} \; J \cdot s$, speed of light $c = 3 \times 10^8 \; m/s$)

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