For a p-n junction the intensity of the electric field is 1×10⁶ V m⁻¹ and the width of the depletion region is 5000 A. The value of the potential barrier is V.

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(3) 0.5. Given: E = 1×10⁶ V m⁻¹ and a depletion width w = 5000 A = 5000×10⁻¹⁰ m = 5×10⁻⁷ m. Idea: inside the depletion layer the field of the uncovered donor and acceptor ions is treated as uniform, so the potential it builds up across the layer is simply V_b = E w. V_b = 1×10⁶ × 5×10⁻⁷ V_b = 0.5 V, which is the order expected for a silicon junction.

For a p-n junction the intensity of the electric field is 1×10⁶ V m⁻¹ and the width of the depletion region is 5000 A. The value of the potential barrier is ________ V.

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