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Electronic Devices question

2022 · 29 Jul · Shift 1 · Q76
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Electronic Devices question

2022 · 29 Jul · Shift 1 · Q76

JEE MainPhysicsElectronic DevicesNumerical+4 / −1
If the potential barrier across a p-n junction is 0.6 V0.6 \mathrm{~V}0.6 V. Then the electric field intensity, in the depletion region having the width of 6×10−6 m6 \times 10^{-6} \mathrm{~m}6×10−6 m, will be ‾×105 N/C\underline{\hspace{2cm}}\times 10^{5} \mathrm{~N} / \mathrm{C}​×105 N/C.
Numerical answer
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Correct answer: 1

  1. For a p-n junction, the electric field intensity across the depletion region is approximated by

E=VdE = \frac{V}{d}E=dV​

where:

  • V=0.6 VV = 0.6\,\text{V}V=0.6V is the potential barrier,
  • d=6×10−6 md = 6 \times 10^{-6}\,\text{m}d=6×10−6m is the depletion width.
  1. Substitute the values:

E=0.66×10−6E = \frac{0.6}{6 \times 10^{-6}}E=6×10−60.6​

  1. Simplify:

E=0.66×106=0.1×106=1×105 N/CE = \frac{0.6}{6} \times 10^{6} = 0.1 \times 10^{6} = 1 \times 10^{5}\,\text{N/C}E=60.6​×106=0.1×106=1×105N/C

  1. The question asks for the value in the form

‾×105 N/C\underline{\hspace{1cm}} \times 10^{5}\,\text{N/C}​×105N/C

So the required integer is:

111

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