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प्रश्न
In a solid the energy level is lying 0.012 eV below Fermi level. What is the probability of this level not being occupied by an electron 27℃?
उत्तर
:-Data :- -EF E=0.012eV ,T=27℃=300K
`K=1.38×10^(-23) J/K= =(1.38×10^(-23))/(1.6×10^(-19 ))=86.25×10^-6 eV/K `
Formula :-f(Ec)= `1/(1+e("^(E_c- E_F))/(kT )`
Calculations:-Total probability = 1
Probability of occupying an energy state + Probability of not occupying the energy state = 1. f(E) + Probability of not occupying the energy state = 1 Probability of not occupying the energy state = 1- f(E)
`"Here" f(E) =1/(1+e((E-E_f))/(kT )) =1/(1+e^((0.012/86.25×10^-6×300))) = 0.386`
Hence ,1-f(E)=1-0.386=0.614
Answer:-Probability of not occupying=0.614
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