A1
^{ ??}
Find the charge $dq$ flowing in time $dt$ through the area $S$ perpendicular to the field.

Using Ohm's law $R = U/I = \rho l/S$, express the resistivity $\rho$ of a semiconductor through $\mu_n$, $\mu_p$, $n$, and $p$.

Using Ohm's law $R = U/I = \rho l/S$, express the resistivity $\rho$ of a semiconductor through $\mu_n$, $\mu_p$, $n$, and $p$.

Both charge carriers participate in a current, so $dq = e n v_n S \, dt + e p v_p S \, dt$, where $v_n = \mathcal{E} \mu_n$ and $v_p = \mathcal{E} \mu_p$. So, on the one hand

\[ dq = \mathcal{E}eS dt (n \mu_n + p \mu_p)\]

and on the other hand

\[ \frac{dq}{dt} = I = \frac{U}{R} = \frac{\mathcal{E} S}{\rho}\]

Answer:
\[ \rho = \frac{1}{\mu_p e p + \mu_n e n}\]

A2
^{ ??}
Which charge carrier is dominant in the sample if the measured voltage is $V_H=+1.25~\text{mV}$? What is the charge carrier concentration in the sample?

Let's describe both cases. The second Newton's law for the particle is

\[ q\vec{\mathcal{E}}-\frac{|q|}{\mu}\vec{v}+q\vec{\mathcal{E}}_\perp + q \vec{v} \times \vec{B} = 0\]

Thus, the case of the stationary motion is described by $\vec{\mathcal{E}}_\perp + \vec{v} \times \vec{B} = 0$.

Since $V_H$ is positive the direction of the electric field is from the bottom to the top. This is the case of negative charge carriers: electrons.

\[\frac{V_H}{W} = \mathcal{E}_\perp, \quad \mathcal{E}_\perp = v_nB = \mu_n \mathcal{E}B= \mu_n \rho_n j B = \frac{IB}{en A}\]

Answer:
The major carriers are electrons and their concentration is

\[ n = \frac{WIB}{AeV_H} = 1.0 \cdot 10^{18}~\text{m}^{-3}\]

\[ n = \frac{WIB}{AeV_H} = 1.0 \cdot 10^{18}~\text{m}^{-3}\]

A3
^{ ??}
A measurement of $V_H=+1.25~\text{mV}$ was made in winter at a temperature of $T=290~\text{K}$ in a laboratory with poor thermoregulation. What value of $V'_H$ would be measured with the same setup in the summer, when the temperature in the lab reaches $T=305~\text{K}$?

We have the electronetruality condition $n-N_A = p$, which has the form

\[ n_i \sqrt{\frac{T}{T_0}} e^{\frac{\Delta E_F}{k_\text{B}T}} - N_A = n_i \sqrt{\frac{T}{T_0}} e^{-\frac{\Delta E_F}{k_\text{B}T}}.\]

Also $n \gg n_i$, then $p \simeq 0$ and $n \simeq N_A$. It means that $V_H$ doesn't depend on the temperature in this case.

Answer:
\[V_H' = V_H = 1.25~\text{mV}\]