Semiconductors

JEE Main level, numerical answer, medium

54 of 209

Question 54 of 209

If the potential barrier across a p−np-n junction is 0.6V0.6V. Then the electric field intensity, in the depletion region having the width of 6×10−6m6\times {10}^{-6}m, will be _____ ×105NC−1\times {10}^{5}N{C}^{-1}

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Answer: 1


Electric field intensity is E=Vd=PotentialbarrieracrossjunctionwidthofdepletionregionE=\dfrac{V}{d}=\dfrac{Potentialbarrieracrossjunction}{widthofdepletionregion}
=0.6V6×10−6m=1×105Vm−1=\dfrac{0.6V}{6\times {10}^{-6}m}=1\times {10}^{5}V{m}^{-1}
=1×105NC−1=1\times {10}^{5}N{C}^{-1}

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