Answer to Question #235203 in Electricity and Magnetism for sonu

Question

Two semiconductor materials have exactly the same properties except that material A has a

bandgap energy of 1.0 eV and material B has a bandgap energy of 1.2 eV. Determine the ratio

of ni, of material A to that of material B for T = 300 K.

Answer 1

We have to use the relation for the holes on the semiconductor as:

$\cfrac{n_{i,A}^2}{n_{i,B}^2}=\large { \cfrac{e^{-\frac{E_g^A}{kT} } }{e^{-\frac{E_g^B}{kT} } } } = e^{({\frac{E_g^B-E_g^A}{kT}})} \\ \text{we use the value for k in eV/K and substitute to find the relation:} \\ \cfrac{n_{i,A}^2}{n_{i,B}^2}=e^{\frac{(1.2-1.0)eV}{(8.617 \times 10^{-5}\,eV/K)(300\,K)} }=e^{7.7366} \\ \implies \cfrac{n_{i,A}}{n_{i,B}}=e^{3.8683}=47.86$

After substitution we find that the relation $\cfrac{n_{i,A}}{n_{i,B}} \text { is equal to } 47.86$.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.