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

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(C) The energy required to create a hole is given by the acceptor level energy: $E = 57 \, meV = 57 	imes 10^{-3} \, eV$.Converting this energy into

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$217100$

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(C) The energy required to create a hole is given by the acceptor level energy: $E = 57 \, meV = 57 	imes 10^{-3} \, eV$.Converting this energy into Joules: $E = 57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} \, J = 9.12 	imes 10^{-21} \, J$.The relationship between energy and wavelength is $E = \frac{hc}{\lambda}$, so $\lambda = \frac{hc}{E}$.Substituting the values: $\lambda = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{9.12 	imes 10^{-21}} \, m$.$\lambda = \frac{19.8 	imes 10^{-26}}{9.12 	imes 10^{-21}} \, m \approx 2.171 	imes 10^{-5} \, m$.Converting meters to $\mathring{A}$s $(\mathring{A})$: $1 \, m = 10^{10} \, \mathring{A}$.$\lambda = 2.171 	imes 10^{-5} 	imes 10^{10} \, \mathring{A} = 217100 \, \mathring{A}$.

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

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