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MHT CET · Physics · Semiconductors

In an n-p-n transistor 200 electrons enter the emitter in \(10^{-8}\) second. If \(1 \%\) electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively \(\left[\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right]\)

  1. A \(2 \times 10^{-10} \mathrm{~A}\) and 49
  2. B \(3.2 \times 10^{-9} \mathrm{~A}\) and 99
  3. C \(1.6 \times 10^{-19} \mathrm{~A}\) and 90
  4. D \(1.7 \times 10^{-11} \mathrm{~A}\) and 70
Verified Solution

Answer & Solution

Correct Answer

(B) \(3.2 \times 10^{-9} \mathrm{~A}\) and 99

Step-by-step Solution

Detailed explanation

\(\mathrm{q}=200 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{t}=10^{-8} \mathrm{~s} \)
\( \therefore \text {Emitter current } \mathrm{I}_{\mathrm{e}}=\frac{\mathrm{q}}{\mathrm{t}}=\frac{3.2 \times 10^{-17}}{10^{-8}}\) \(=3.2 \times 10^{-9} \mathrm{~A} \)
\( \mathrm{I}_{\mathrm{b}}=\frac{1}{100} \cdot \mathrm{I}_{\mathrm{e}} \)
\( \mathrm{I}_{\mathrm{c}}=\frac{99}{100} \cdot \mathrm{I}_{\mathrm{e}}\)
\(\therefore\) Current amplification factor \(\beta=\frac{I_c}{I_b}=99\)