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In NPN transistor, $10^{10}$ electrons enter in emitter region in $10^{- 6}$ s. If 2% electrons are lost in base region, then collector current and current amplification factor $(β)$ respectively are

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1.57 mA, 49

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detailed solution

Correct option is A

Ie=1010×1.6×10−19×110−6=1.6 mA ∵I=Qtsince 2% electrons are absorbed by base, hence 98% electrons reaches the collector, i.e., α=0.98⇒Ic=αIe=0.98×1.6=1.568 mA ≈1.57 mA

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In NPN transistor, 1010 electrons enter in emitter region in 10-6 s. If 2% electrons are lost in base region, then collector current and current amplification factor β respectively are

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In NPN transistor, $10^{10}$ electrons enter in emitter region in $10^{- 6}$ s. If 2% electrons are lost in base region, then collector current and current amplification factor $(β)$ respectively are