Consider a n-type semiconductor whose EV=0, EC=2.0 eV and ED=1.98 eV. The correct statement among the… · ABC26PH0191
Subject: Physical Chemistry · Chapter: Solid State Chemistry · Topic: Solid State Chemistry – General · Exam: CSIR NET 2011 · Marks: 2 · Difficulty: Easy
Consider a n-type semiconductor whose $E_{V}=0$, $E_{C}=2.0\ \mathrm{eV}$ and $E_{D}=1.98\ \mathrm{eV}$. The correct statement among the following is
(a)$E_{F}=1\ \mathrm{eV}$ and is independent of T
(b)$\mathrm{E}_{\mathrm{F}}=\mathrm{1 . 99 ~ eV}$ and remains independent of T
(c)$\mathrm{E}_{\mathrm{F}}=\mathrm{1 . 99 ~ eV}$ and increases towards 2.0 eV with increase of T
(d)$\mathrm{E}_{\mathrm{F}}=1.99 \mathrm{eV}$ and decreases with increase of T.
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Key was missing. For an n-type semiconductor at low T, E_F lies midway between E_D and E_C: (1.98 + 2.0)/2 = 1.99 eV; as T rises the donors ionise and intrinsic carriers dominate, so E_F falls toward the mid-gap value (1.0 eV). Hence (d).
Explanation
At low temperature the Fermi level of an n-type semiconductor lies midway between the donor level and the conduction-band edge: $E_F=\dfrac{E_D+E_C}{2}=\dfrac{1.98+2.0}{2}=1.99\ \mathrm{eV}$.As T increases, the donors become fully ionised and intrinsic excitation across the gap takes over, so $E_F$ moves down toward the intrinsic (mid-gap) value $\dfrac{E_V+E_C}{2}=1.0\ \mathrm{eV}$, i.e. $E_F$ decreases with increase of T.Answer: (d)