Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: Dec 2014

$$E_f = 5.5ev = 5.5 × 1.6 × 10^{-19}m = 8.8 × 10^{-19}$$ $$T = 300k$$ $$F(E) = 0.9$$

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

$ \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}$

$-0.955 × 414 = E - 8.8 × 10^4$

$E = 8.8 × 10^4 - 0.955 × 414$

$E = 8.7604 × 10^4v$