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MHT CET · Physics · Semiconductors

In an NPN transistor \(10^{10}\) electrons enter the emitter in \(10^{-6} \mathrm{~s}\) and \(2 \%\) electrons recombine with holes in base. The current ratios ' \(\alpha\) ' and ' \(\beta\) ' of a transistor are respectively (nearly)

  1. A \(0 \cdot 98,49\)
  2. B \(49,0.98\)
  3. C \(0 \cdot 49,98\)
  4. D \(98,0.49\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0 \cdot 98,49\)

Step-by-step Solution

Detailed explanation

\(\mathrm{I}_{\mathrm{e}}=\frac{\mathrm{n}_{\mathrm{e}} \times \mathrm{e}}{\mathrm{t}} \text { and } \mathrm{I}_{\mathrm{c}}=\frac{\mathrm{n}_{\mathrm{c}} \times \mathrm{e}}{\mathrm{t}}\)
From the data given,
\(\begin{aligned}
I_c & =\frac{(98 / 100) n_e \times e}{t}=\frac{98}{100} I_c \\
\alpha & =\frac{I_c}{I_e}=\frac{98 I_e}{100 I_e}=0.98
\end{aligned}\)
\(\therefore \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=\frac{0.98}{0.02}=49\)