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Electronic Devices question

2021 · 25 Jul · Shift 2 · Q71
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Electronic Devices question

2021 · 25 Jul · Shift 2 · Q71

JEE MainPhysicsElectronic DevicesNumerical+4 / −1
In a semiconductor, the number density of intrinsic charge carries at 27 ∘^\circ∘ C is 1.5 ×\times× 1016/m3. If the semiconductor is doped with impurity atom, the hole density increases to 4.5 ×\times× 1022/m3. The electron density in the doped semiconductor is ‾\underline{\hspace{2cm}}​×\times× 109/m3.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Use mass action law for semiconductors

For any semiconductor in thermal equilibrium,

np=ni2np = n_i^2np=ni2​

where:

  • nnn = electron density in doped semiconductor
  • ppp = hole density in doped semiconductor
  • nin_ini​ = intrinsic carrier concentration
  1. Given data

ni=1.5×1016 m−3n_i = 1.5 \times 10^{16}\, \text{m}^{-3}ni​=1.5×1016m−3

p=4.5×1022 m−3p = 4.5 \times 10^{22}\, \text{m}^{-3}p=4.5×1022m−3

We need to find nnn.

  1. Substitute into the relation

n=ni2pn = \frac{n_i^2}{p}n=pni2​​

First compute ni2n_i^2ni2​:

ni2=(1.5×1016)2=2.25×1032n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32}ni2​=(1.5×1016)2=2.25×1032

Now,

n=2.25×10324.5×1022n = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}}n=4.5×10222.25×1032​

  1. Simplify

n=2.254.5×1032−22n = \frac{2.25}{4.5} \times 10^{32-22}n=4.52.25​×1032−22

n=0.5×1010n = 0.5 \times 10^{10}n=0.5×1010

n=5×109 m−3n = 5 \times 10^9\, \text{m}^{-3}n=5×109m−3

  1. Required integer

The question asks for:

‾×109/m3\underline{\hspace{2cm}} \times 10^9/\text{m}^3​×109/m3

So the blank is:

555

  1. Comparison with stored answer

Stored correct answer = 555

This matches our derived answer.

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