### Exercise 1: Extracting quantities from basic Hall measurements
1. The Hall voltage is measured across the sample width. Hence,
$$
V_H = -\int_{0}^{W} E_ydy
$$
$$
V_H = V(W)-V(0) = -\int_{0}^{W} E_ydy = -E_y W
$$
where $E_y = v_xB$, which we obtain from the steady-state Drude equation. The (surface) current density is $I = j_x W$. Thus $R_{xy} = \frac{v_xB}{j_x} = \frac{-B}{ne}$,
so it does not depend on the sample geometry.
where $E_y = \rho_{yx}j_x = \frac{-B}{ne}j_x = v_xB$, which we obtain from the steady-state Drude equation. The bias current is $I_x = j_x W$, with $j_x$ the surface current density. So we get $R_{xy} = V_H/I_x = \frac{B}{ne}$.
so it does not depend on the sample geometry.
2. If hall resistance and magnetic field are known, the charge density is calculated from $R_{xy} = -\frac{B}{ne}$.
2. If the hall resistance and the magnetic field are known, we extract the charge density using $R_{xy} = -\frac{B}{ne}$.
As $V_x = -\frac{I_x}{ne}B$, a stronger field yields a larger Hall voltage, making it easier to measure. Likewise, a lower charge density yields a larger Hall voltage (for the same value of the bias current), making it easier to measure.
As $V_x = -\frac{I_x}{ne}B$, a stronger field yields a larger Hall voltage, making it easier to measure. Likewise, a lower charge density yields a larger Hall voltage, making it easier to measure.
3. The resistance is
$$
R_{xx} = \frac{\rho_{xx}L}{W}
$$
$$
R_{xx} = \frac{\rho_{xx}L}{W}
$$
where $\rho_{xx} = \frac{m_e}{ne^2\tau}$. Therefore, knowing the charge density $n$ and assuming the current is carried by electrons, we can extract the scattering time ($\tau$). $R_{xx}$ depends on the sample geometry, but $\tau$, and $n$ do not.
where $\rho_{xx} = \frac{m_e}{ne^2\tau}$ is the longitudinal Drude resistivity. Therefore, knowing the electron density $n$, we can extract the scattering time ($\tau$). We observe that $R_{xx}$ depends on the sample geometry ($L$ and $W$, which is in contrast with the Hall resistance $R_{xy}$.
<!---
#### Question 4.
...
...
@@ -38,11 +38,7 @@ See 3_drude_model.md
### Exercise 2: Temperature dependence of resistance in the Drude model
1. Find electron density from $n_e = \frac{ZnN_A}{W} $
where *Z* is valence of copper atom, *n* is density, $N_A$ is Avogadro constant and *W* is atomic weight. Use $\rho$ from the lecture notes to calculate scattering time.
($ \tau = 2.57 \cdot 10^{-14} s$)
1. We first find the electron density using $n_e = \frac{ZnN_A}{W}$, where $Z=1$ is the number of free electrons per copper atom, *n* is the mass density of copper, $N_A$ is Avogadro's constant, and *W* is the atomic weight of copper. Then, we find the scattering time $\tau = 2.57 \cdot 10^{-14}$ s from the longitudinal Drude resistivity $\rho = \frac{m_e}{n_e e^2\tau}$.
2. $\lambda = \langle v \rangle\tau$ ($\lambda =3 nm $)
3. Scattering time $\tau \propto \frac{1}{\sqrt{T}}$; $\rho \propto \sqrt{T}$
4. In general, $\rho \propto T$ as the phonons in the system scales linearly with T (remember high temperature limit of Bose-Einstein factor becomes $\frac{kT}{\hbar\omega}$ leading to $\rho \propto T$). Inability to explain this linear dependence is a failure of the Drude model.
...
...
@@ -52,13 +48,13 @@ See 3_drude_model.md
1. $\rho_{xx}$ is independent of B and $\rho_{xy} \propto B$
3. This describes a [Lorentzian](https://en.wikipedia.org/wiki/Spectral_line_shape#Lorentzian).
...
...
@@ -69,9 +65,9 @@ See 3_drude_model.md
### Exercise 4. Positve and negative charge carriers
1.
$$
$\bf J$ = -n_e e \mathbf{ v_e} + n_h e \mathbf{ v_h}
$$
$$
\mathbf{J} = -n_e e \mathbf{ v_e} + n_h e \mathbf{ v_h}
$$
2. Write down the Drude equation for the positive and negative charge carriers seperately, $ \mu_e = \frac{e\tau_e}{m_e}$ and $ \mu_h = \frac{e\tau_h}{m_h}$
The sign is determined by a weighted average of the charge density and the mobilty of the positive and negative charge carriers. Therefore, a measurement of the Hall voltage reveals the dominant charge carrier
