Fermi Energy for silver is 5-5eV. Find out energy for which the probability of occupancy at 300 K is 0.9.

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written 8.7 years ago by

teamques10 ★ 69k

• modified 8.7 years ago

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: Dec 2014

engineering physics

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written 8.7 years ago by

teamques10 ★ 69k

• modified 8.7 years ago

$E_f = 5.5ev = 5.5 × 1.6 × 10^{-19}m = 8.8 × 10^{-19}$

$T = 300k$

$F(E) = 0.9$

$F(E) = \frac{1}{1 + exp^{\frac{E - E_f}{KT}}}$

$\frac{1}{F(E)} = 1 + exp^{\frac{E - E_f}{KT}}$

$\frac{1}{8.8 × 10^{-19}} = 1 + exp^{\frac{E - E_f}{KT}}$

$Log (1.11 - 1) = log exp^{\frac{E - 5.5 × 1.6 × 10^{-19}}{1.38 × 10^{-23} × 300}}$

$0.955 = \frac{E - 5.5 × 1.6 × 10^{-19}}{1.38 × 10^{-23} × 300}$

$-0.955 × 414 = E - 8.8 × 10^4$

$E = 8.8 × 10^4 - 0.955 × 414$

$E = 8.7604 × 10^4v$

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