Fermi level with 10¹⁰ Sb then 2×10¹⁵ B

Take $T = 300$ K, $kT = 0.02586$ eV and $n_i = 1.45\times10^{10}$ cm⁻³ for Si. Fermi level positions are measured from the intrinsic Fermi level $E_{Fi}$.

Part 1: Only $N_d = 10^{10}$ cm⁻³ antimony

Here $N_d$ is smaller than $n_i$, so we cannot use $n \approx N_d$. Use charge neutrality $n = N_d + p$ with $np = n_i^2$:

$$\begin{aligned} n^2 - N_d n - n_i^2 &= 0 \\ n &= \frac{N_d}{2} + \sqrt{\left(\frac{N_d}{2}\right)^2 + n_i^2} \\ &= 0.5\times10^{10} + \sqrt{(0.5\times10^{10})^2 + (1.45\times10^{10})^2} \\ &= 2.03\times10^{10}\ \text{cm}^{-3} \end{aligned}$$

Since $n = n_i \exp[(E_F - E_{Fi})/kT]$:

$$E_F - E_{Fi} = kT\ln\frac{n}{n_i} = 0.02586\ln\frac{2.03\times10^{10}}{1.45\times10^{10}} = 0.0087\ \text{eV}$$

The Fermi level is only about 8.7 meV above $E_{Fi}$; such light doping makes the wafer almost intrinsic.

Part 2: Further doping with $N_a = 2\times10^{15}$ cm⁻³ boron

Now $N_a > N_d$, so the sample becomes p-type (compensated). Net acceptor concentration:

$$N_a - N_d = 2\times10^{15} - 10^{10} \approx 2\times10^{15}\ \text{cm}^{-3} \gg n_i$$

So $p \approx 2\times10^{15}$ cm⁻³ and $n = n_i^2/p \approx 1.05\times10^{5}$ cm⁻³.

$$\begin{aligned} E_{Fi} - E_F &= kT\ln\frac{p}{n_i} = 0.02586 \ln\frac{2\times10^{15}}{1.45\times10^{10}} \\ &= 0.02586 \times 11.83 = 0.306\ \text{eV} \end{aligned}$$

 Ec ---------------------------
     EF (part 1) 0.0087 eV above Ei
 Ei - - - - - - - - - - - - - -
                 | 0.306 eV
 EF (part 2) ----+-------------
 Ev ---------------------------

Answer: With $10^{10}$ cm⁻³ Sb, $E_F$ is 0.0087 eV (8.7 meV) above $E_{Fi}$. After adding $2\times10^{15}$ cm⁻³ B, the sample is p-type and $E_F$ shifts to 0.306 eV below $E_{Fi}$, a total shift of about 0.315 eV.

Note: if the textbook version of this problem ($10^{16}$ Sb, then $2\times10^{17}$ B) is intended, the same method gives $E_F - E_{Fi} = 0.348$ eV, then $E_{Fi} - E_F = 0.424$ eV.