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Answer (Detailed Solution Below)

Option 1 : $2.25 \times 10^{4}~\text{cm}^{-3}$ / $2.25 \times 10^{4}~\text{सेमी}^{-3}$

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Detailed Solution

सही उत्तर $2.25 \times 10^{4}~\text{cm}^{-3}$ / $2.25 \times 10^{4}~\text{सेमी}^{-3}$ है।

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Question:

In a silicon crystal doped with $10^{16}~\text{cm}^{-3}$ donor atoms at $300$ K ($n_{i} = 1.5 \times 10^{10}~\text{cm}^{-3}$), the equilibrium minority-carrier (hole) concentration is / $300$ केल्विन पर $10^{16}~\text{सेमी}^{-3}$ दाता परमाणुओं से अपमिश्रित सिलिकॉन क्रिस्टल में साम्य अल्पसंख्यक-वाहक (होल) सांद्रता है

1.
$2.25 \times 10^{4}~\text{cm}^{-3}$ / $2.25 \times 10^{4}~\text{सेमी}^{-3}$
2.
$1.5 \times 10^{10}~\text{cm}^{-3}$ / $1.5 \times 10^{10}~\text{सेमी}^{-3}$
3.
$2.25 \times 10^{10}~\text{cm}^{-3}$ / $2.25 \times 10^{10}~\text{सेमी}^{-3}$
4.
$2.25 \times 10^{16}~\text{cm}^{-3}$ / $2.25 \times 10^{16}~\text{सेमी}^{-3}$

Correct Answer: $2.25 \times 10^{4}~\text{cm}^{-3}$ / $2.25 \times 10^{4}~\text{सेमी}^{-3}$

By the mass-action law $np = n_{i}^{2}$, so $p = n_{i}^{2}/N_{d} = (1.5 \times 10^{10})^{2}/10^{16} = 2.25 \times 10^{4}~\text{cm}^{-3}$. / द्रव्यमान-क्रिया नियम $np = n_{i}^{2}$ से $p = n_{i}^{2}/N_{d} = (1.5 \times 10^{10})^{2}/10^{16} = 2.25 \times 10^{4}~\text{सेमी}^{-3}$।

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