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1 year ago

Find the electron and hole concentrations and fermi level in silicon at 300 k for boron

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To find the electron and hole concentrations and fermi levels in silicon at 300 k for boron, fermi level is 9.3 x $10^4 cm^-3$ .

<h3>Calculation and Explanation</h3>

Ionization energy for boron in Si, 0.045 eV

Fermi level in silicon,

300 K at $10^{15} cm^{-3}$ (Boron atoms)

All boron impurities are ionized, 300K

Electron, hole concentrations and Fermi levels is found out,

The value of Na = $10^{15} cm^{-3}$

np = $\frac{ni^2}{nA}$

np = $\frac{(9.65 x 10^9)^2}{10^15}$

np= 9.3 x $10^4 cm^-3$

Fermi level is calculated,

Ef - Ev = kT ln(NV/ND)

value of kT = 0.0259 eV (300° K)

So, substitution the values

Ef - Ev = kT ln(NV/ND)

Ef - Ev = 0.0259ln (2.66 x $10^{19} / 10^{15}$ )

Ef - Ev = 0.0259ln (26600)

Ef - Ev = 0.0259 x 10.18

Ef - Ev = 0.263 eV

What is ionization energy?

Ionization energy, also known as ionization energy (IE) or ionization energy (British English spelling), is the minimal amount of energy needed to free the most loosely bonded electron from an isolated gaseous atom, positive ion, or molecule in physics and chemistry.

To know more about ionization energy, visit:

brainly.com/question/16243729

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