In a P-type semiconductor, the acceptor energ | vedclass

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(C) The energy required to create a hole is equal to the acceptor energy level $E_A = 57 \, meV = 57 	imes 10^{-3} \, eV$. Converting this energy int

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$217100$

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(C) The energy required to create a hole is equal to the acceptor energy level $E_A = 57 \, meV = 57 	imes 10^{-3} \, eV$. Converting this energy into Joules: $E_A = 57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} \, J = 91.2 	imes 10^{-22} \, J$. The relation between energy and wavelength is $E = \frac{hc}{\lambda }$. Therefore, $\lambda = \frac{hc}{E_A} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19}} \, m$. $\lambda = \frac{19.8 	imes 10^{-26}}{91.2 	imes 10^{-22}} \, m = 0.2171 	imes 10^{-4} \, m$. Converting to $\mathring{A}$s: $\lambda = 0.2171 	imes 10^{-4} 	imes 10^{10} \, \mathring{A} = 217100 \, \mathring{A}$.

In a $P$-type semiconductor, the acceptor energy level $E_A$ is located $57 \, meV$ above the valence band. The maximum wavelength required to create a hole is ....... $\mathring{A}$. (Given: $h = 6.6 \times 10^{-34} \, J \, s$, $c = 3 \times 10^8 \, m/s$, $1 \, eV = 1.6 \times 10^{-19} \, J$)

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