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A p type semiconductor has acceptor levels 57 meV above the valance band. The maximum wavelength of light required to create hole is (Planck constant \(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}\) )
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Verified Answer
The correct answer is:
\(217105 Å\)
Given:
\(\begin{aligned}
& \mathrm{E}=57 \mathrm{meV}=57 \times 10^{-3} \times 1.6 \times 10^{-19} \mathrm{~J} \\
& \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \\
& \lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{57 \times 1.6 \times 10^{-22}} \\
& \lambda=\frac{6.6 \times 3 \times 10^{-26}}{57 \times 1.6 \times 10^{-22}} \\
& \lambda=0.2171052 \times 10^{-4} \mathrm{~m} \\
& \lambda=217105.2 \times 10^{-10} \mathrm{~m} \\
& \lambda=217105 Å \text { (Approx.) }
\end{aligned}\)
\(\begin{aligned}
& \mathrm{E}=57 \mathrm{meV}=57 \times 10^{-3} \times 1.6 \times 10^{-19} \mathrm{~J} \\
& \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \\
& \lambda=\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{57 \times 1.6 \times 10^{-22}} \\
& \lambda=\frac{6.6 \times 3 \times 10^{-26}}{57 \times 1.6 \times 10^{-22}} \\
& \lambda=0.2171052 \times 10^{-4} \mathrm{~m} \\
& \lambda=217105.2 \times 10^{-10} \mathrm{~m} \\
& \lambda=217105 Å \text { (Approx.) }
\end{aligned}\)
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