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GUJCET · Physics · Semiconductor Electronic: Material, Devices And Simple Circuits
Suppose a pure \(\mathrm{S}_{i}\) crystal has \(5 \times 10^{28}\) atoms \(\mathrm{m}^{-3}\). It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that \(n_{i}=1.5 \times 10^{16} \mathrm{~m}^{-3}\)
- A \(5.4 \times 10^{-9} \mathrm{~m}^{-3}\)
- B \(5.4 \times 10^{9} \mathrm{~m}^{-3}\)
- C \(4.5 \times 10^{-9} \mathrm{~m}^{-3}\)
- D \(4.5 \times 10^{9} \mathrm{~m}^{-3}\)
Answer & Solution
Correct Answer
(D) \(4.5 \times 10^{9} \mathrm{~m}^{-3}\)
Step-by-step Solution
Detailed explanation
D.\(4.5 \times 10^{9} \mathrm{~m}^{-3}\)
No. of As atoms in \(10^{6} \mathrm{~S}_{i}=1\)
\(\therefore \quad\) No. of As atoms in \(5 \times 10^{28} \mathrm{~S}_{i}=n_{e}\)
\(\therefore n_{e}=\frac{5 \times 10^{28}}{10^{6}}\) \(=5 \times 10^{22} \mathrm{~m}^{-3}\)
Now \(n_{i}^{2}=n_{e} n_{h}\)
\(\therefore \quad n_{h}=\frac{n_{i}^{2}}{n_{e}}\)
\(\therefore \quad n_{h}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{5 \times 10^{22}}\)
\(\therefore \quad n_{h}=4.5 \times 10^{9} \mathrm{~m}^{-3}\)
No. of As atoms in \(10^{6} \mathrm{~S}_{i}=1\)
\(\therefore \quad\) No. of As atoms in \(5 \times 10^{28} \mathrm{~S}_{i}=n_{e}\)
\(\therefore n_{e}=\frac{5 \times 10^{28}}{10^{6}}\) \(=5 \times 10^{22} \mathrm{~m}^{-3}\)
Now \(n_{i}^{2}=n_{e} n_{h}\)
\(\therefore \quad n_{h}=\frac{n_{i}^{2}}{n_{e}}\)
\(\therefore \quad n_{h}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{5 \times 10^{22}}\)
\(\therefore \quad n_{h}=4.5 \times 10^{9} \mathrm{~m}^{-3}\)
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