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In a p–n junction a potential barrier of 300 meV exists across the junction. A hole with a kinetic energy of 400 meV approaches the junction. The K.E of the hole when it crosses the junction (i) from p to n–side and (ii) from n to p–side will be (1) 100 eV, 700 eV(2) 100 eV, 100 eV(3) 700 eV, 700 eV(4) 700 eV, 100 eV |
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Answer» Correct option: (1) 100 eV, 700 eV Explanation: The p–region is at a lower potential than n–region in depletion layer when hole crosses depletion layer from n– to p–side; it (hole) is accelerated. Therefore K = Kinetic energy of hole = Ki + |e| V = (400 + 300) meV = 700 meV When hole crosses depletion layer from p– to n–side, it is retarded. Therefore K’ = Ke – |e| V = (400–300) meV = 100 meV |
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