From 3faec748dde1db179f019a9a6594a1f8652394a4 Mon Sep 17 00:00:00 2001
From: Bowy La Riviere <b.m.lariviere@student.tudelft.nl>
Date: Thu, 18 Feb 2021 22:25:12 +0000
Subject: [PATCH] Updates some answers
---
src/4_sommerfeld_model_solutions.md | 29 +++++++++++++++++------------
1 file changed, 17 insertions(+), 12 deletions(-)
diff --git a/src/4_sommerfeld_model_solutions.md b/src/4_sommerfeld_model_solutions.md
index 0538b1bd..85dd28bd 100644
--- a/src/4_sommerfeld_model_solutions.md
+++ b/src/4_sommerfeld_model_solutions.md
@@ -40,21 +40,21 @@ Comparing total and free electron density, only few electrons are available for
2.
$$
-g_{1D}(k)dk = \left(\frac{L}{2\pi}\right) 2 \mathrm{d} k
+g_{1D}(k)\textrm{d} k = \left(\frac{L}{2\pi}\right) 2 \mathrm{d} k
$$
The factor 2 is due to positive and negative $k$-values having equal enery
$$
-g_{2D}(k)dk = \left(\frac{L}{2\pi}\right)^2 2\pi k \mathrm{d} k
+g_{2D}(k)\textrm{d} k = \left(\frac{L}{2\pi}\right)^2 2\pi k \mathrm{d} k
$$
$$
-g_{3D}(k)dk = \left(\frac{L}{2\pi}\right)^3 4\pi k^2 \mathrm{d} k
+g_{3D}(k)\textrm{d} k = \left(\frac{L}{2\pi}\right)^3 4\pi k^2 \mathrm{d} k
$$
3.
$$
\begin{align}
-g(k)dk &= \left(\frac{L}{2\pi}\right)^n S_{n-1}(k)dk \\
-&= \left(\frac{L}{2\pi}\right)^n \frac{2\pi^{\frac{n}{2}}k^{n-1}}{\Gamma(\frac{n}{2})}dk
+g(k)\textrm{d} k &= \left(\frac{L}{2\pi}\right)^n S_{n-1}(k)\textrm{d} k \\
+&= \left(\frac{L}{2\pi}\right)^n \frac{2\pi^{\frac{n}{2}}k^{n-1}}{\Gamma(\frac{n}{2})}\textrm{d} k
\end{align}
$$
@@ -62,10 +62,10 @@ $$
5.
$$
-g(\varepsilon)d\varepsilon = 2_s g(k) dk
+g(\varepsilon)\textrm{d} \varepsilon = 2_s g(k) \textrm{d} k
$$
$$
-g(\varepsilon)=2_sg(k(\varepsilon))\frac{dk}{d\varepsilon}
+g(\varepsilon)=2_sg(k(\varepsilon))\frac{\textrm{d} k}{\textrm{d} \varepsilon}
$$
$$
g(\varepsilon) = \frac{2_s}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m}{2\pi}}\right)^n (\varepsilon)^{\frac{n}{2}-1}
@@ -73,24 +73,29 @@ $$
6.
$$
-N = \int_{0}^{\infty}g(\varepsilon)n_F(\beta(\varepsilon-\mu))d\varepsilon = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m}{2\pi}}\right)^n\int_{0}^{\infty}\frac{(\varepsilon)^{\frac{n}{2}-1}}{e^{\frac{\varepsilon-\mu}{k_BT}}+1}d\varepsilon
+N = \int_{0}^{\infty}g(\varepsilon)n_F(\beta(\varepsilon-\mu))\textrm{d} \varepsilon = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m}{2\pi}}\right)^n\int_{0}^{\infty}\frac{(\varepsilon)^{\frac{n}{2}-1}}{e^{\frac{\varepsilon-\mu}{k_BT}}+1}\textrm{d} \varepsilon
$$
-Total energy: $E = \int_{0}^{\infty} g(\varepsilon) n_{F}(\beta (\varepsilon - \mu)) \varepsilon d\varepsilon $
+Total energy: $E = \int_{0}^{\infty} g(\varepsilon) n_{F}(\beta (\varepsilon - \mu)) \varepsilon \textrm{d} \varepsilon $
### Exercise 3: a hypothetical material
1.
$$
-E = \int_{0}^{\infty}\varepsilon g(\varepsilon)f(\varepsilon)d\varepsilon = 2.10^{10}\int_{0}^{\infty}\frac{\varepsilon^{\frac{3}{2}}}{e^\frac{\varepsilon-5.2}{k_BT}+1}d\varepsilon
+E = \int_{0}^{\infty}\varepsilon g(\varepsilon) n_{F}(\beta (\varepsilon - \mu)) \textrm{d} \varepsilon = 2.10^{10}eV^{-\frac{3}{2}} \int_{0}^{\infty}\frac{\varepsilon^{\frac{3}{2}}}{e^\frac{\varepsilon-5.2}{k_BT}+1} \textrm{d} \varepsilon
$$
2.
-Substitute T=0 in the integral expression for total energy to find the ground state energy.
+$$
+E = \frac{4}{5} (5.2)^{\frac{5/2}} 10^{10} eV
+$$
3.
$$
-\Delta E = \frac{\pi^2}{6}(k_B T)^2\frac{\partial}{\partial \varepsilon}\left(\varepsilon g(\varepsilon)\right)\bigg|_{\varepsilon=\varepsilon _F}
+\begin{align}
+E(T)-E(T=0) &= \frac{\pi^2}{6}(k_B T)^2\frac{\partial}{\partial \varepsilon}\left(\varepsilon g(\varepsilon)\right)\bigg|_{\varepsilon=\varepsilon _F}\\
+&\approx 8.356 10^8 eV
+\end{align}
$$
5.
--
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