固体物理代写|Solid-state physics代考|高质保分固体物理方程作业

Problem 1.5s
The sphere and the probability distribution have both inversion and rotation symmetry; the first implies $\langle x\rangle=\langle y\rangle=\langle z\rangle=0$ and the second in combination with the first implies
$$
\Delta x^2=\left\langle x^2\right\rangle=\Delta y^2=\left\langle y^2\right\rangle=\Delta z^2=\left\langle z^2\right\rangle=\frac{1}{3}\left\langle r^2\right\rangle
$$
Hence,
$$
\left\langle\varepsilon_k\right\rangle=\frac{1}{2 m} 3 \Delta p_x^2 \geq \frac{3}{2 m} \frac{\hbar^2}{4} \frac{1}{\Delta x^2}=\frac{9 \hbar^2}{8 m\left\langle r^2\right\rangle}
$$
For a uniform probability, within the sphere of radius $r_0$ and volume $V=$ $(4 \pi / 3) r_0^3,\left\langle r^2\right\rangle=(3 / 5) r_0^2=(3 / 5)(3 / 4 \pi)^{2 / 3} V^{2 / 3}=0.2309 V^{2 / 3}$. Thus $\left\langle\varepsilon_k\right\rangle=$ $4.87 \hbar^2 / m V^{2 / 3}$.

Problem 1.9s
The quantity $\lambda_m$ must depend on:
(a) $\hbar$, since black body’s radiation is of quantum nature
(b) c, since it is an electromagnetic phenomenon
(c) $k_{\mathrm{B}} T$, since $T$ is the only parameter in the spectral distribution of this radiation; furthermore, absolute temperature is naturally associated with Boltzmann’s constant, $k_{\mathrm{B}}$, as a product $k_{\mathrm{B}} T$ with dimensions of energy
Out of $\hbar, c, k_{\mathrm{B}} T$, there is only one combination with dimensions of length $\hbar c / k_{\mathrm{B}} T$ (remember that $\hbar c$ has dimensions of energy times length). Hence,
$$
\lambda_{\mathrm{m}}=c_1 \frac{\hbar c}{k_{\mathrm{B}} T}
$$
where the numerical constant $c_1=1.2655$