Physics 315, Solution to problem set 7 2004

Q3 1993 final exam
(a). When the stresses act normal to the $x-z$ and $y-z$ planes Hooke's law reads

$$\left[\begin{array}{c} -P\\ -P\\ 0\\ 0\\ 0\\ 0\\ \end{a... ...c} e_1\\ e_2\\ e_3\\ e_4\\ e_5\\ e_6\\ \end{array}\right]$$

The solution to these equations is

$$e_1=e_2=-\frac{(\lambda+2\mu)P}{2\mu(3\lambda+2\mu)};\;\;e_3= \frac{\lambda P}{\mu(3\lambda+2\mu)};\;\;e_4=e_5=e_6=0$$

(b). Let the side of the cube be $a$. The sum of the components of the force $-Pa^2$ acting on the the $x-z$ or $y-z$ planes is $-2Pa^2/\sqrt{2}$.The area of a (110) surface is $\sqrt{2}a^2$. The normal stress on the (110) surface is thus $-P$. This result should be expected since the material is isotropic.
Q1 1994 final
a:
The nearest neighbor distance for the $bcc$ structure is $r_{nn}=a\sqrt{3}/2$, where $a$ is the side of the unit cube. We can calculate $a$ from the density $\rho=m/v_c$ using $v_c=a^3/2$ and that the mass $m$ of a sodium atom is $m=M/N_A$, where $M$ is the mass of one mol and $N_A$ is Avogadro's number. Putting the terms together we find

$$r_{nn}=\frac{\sqrt{3}}{2}\left(\frac{2M}{N_A\rho}\right) ^{1/... ...99\times 10^{-3}}{6.0225\times 10^{23}\times 970}\right) ^{1/3}$$

or

$$r_{nn}=3.71\times 10^{-10} m$$

b: The reciprocal of the $bcc$ lattice is $fcc$. We choose the following direct lattice vectors

$${\bf a}=\frac{a}{2}(-\hat{\bf x}+\hat{\bf y}+\hat{\bf z});\; ... ...});\; {\bf c}=\frac{a}{2}(+\hat{\bf x}+\hat{\bf y}-\hat{\bf z})$$

We can choose the primitive reciprocal lattice vectors

$${\bf A}=\frac{2\pi}{a}(\hat{\bf y}+\hat{\bf z});\; {\bf B}=\f... ...{\bf z});\; {\bf C}=\frac{2\pi}{a}(\hat{\bf x}+\hat{\bf y});\;$$

c:
The scattering angles are given by

$$\sin\theta=\frac{n\lambda}{2d}=\frac{n\lambda\vert G\vert} {4\pi}$$

where $\vert G\vert$ is the length of a reciprocal lattice vector. We have $a=2 r_{nn}/\sqrt{3}=4.285$Å. Since $\sin\theta <1$ and $n\geq=1$ we need to consider the lattice vectors for which

$$\vert G\vert<\frac{2\pi}{a}\frac{2a}{\lambda}= \frac{2\pi}{a}5.56565$$

The lengths of the reciprocal lattice vectors can be written on the form

$$\vert G\vert=\frac{2\pi\sqrt{d}}{a}$$

where $d=h^2+k^2+l^2$ and $h,k,l$ are either all even or two of them are odd and one even. The values of $d$ which are smaller than $5.56565^2 =30.97$ are

$$d=2,4,6,8,10,12,16,18,20,24,26$$

each of these $d-$values gives rise to scattering at the angle $\theta=\sin^{-1}\sqrt{d}/5.56565$.
Q 2 1994 final
a:
The relaxation time is given by

$$\tau=\frac{\sigma m_e}{n e^2}=\frac{\sigma m_e v_c}{e^2} =2.93\times 10^{-14} s$$

using the provided numbers, remembering to convert to MKS units!
b:

$$v_{drift}=\frac{e E \tau}{m_e}=\frac{eJ\tau}{\sigma m_e} =0.246\times 10^{-4} ms^{-1}$$

c:
If the cross section of the wire is $A$ and its length $L$ the current is $I=jA$ and the heat dissipated is given by

$$RI^2=\frac{LAj^2}{\sigma}$$

The heat capacity is $3Nk_B=3LAk_B/v_c$ giving for the rate of temperature rise

$$\dot{T}=\frac{v_cj^2}{3\sigma k_B}=0.45\times 10^{-4} Ks^{-1}$$

Q2 1996 final
a: The equations of motion for the atoms in the $n$'th cell can be written

$$m\ddot{u}_1(n)=K_1[u_2(n)-u_1(n)]+K_2[u_2(n-1)-u_1(n)]$$

$$m\ddot{u}_2(n)=K_1[u_1(n)-u_2(n)]+K_2[u_1(n+1)-u_2(n)]$$

b: Substitution of the trial solution yields

$$-m\omega^2e_1=K_1(e_2-e_1)+K_2e_2\exp(-ika)-e_1$$

$$-m\omega^2e_2=K_1(e_1-e_2)+K_2e_1\exp(ika)-e_2$$

For the equations to have a solution the determinant

$$\left\vert\begin{array}{cc} -K_1-K_2+m\omega^2&K_1+K_2\exp(-i... ...K_1+K_2\exp(ika)&-K_1-K_2+m\omega^2\\ \end{array}\right\vert=0$$

The eigenvalues are

$$m\omega^2=\sqrt{K_1+K_2\pm\sqrt{K_1^2+K_2^2+2K_1K_2\cos(ka)}}$$

c: In the special case $ka=\pi$ the frequencies are

$$\omega_1=\sqrt{\frac{2K_1}{m}}$$

$$\omega_2=\sqrt{\frac{2K_2}{m}}$$

d: In the first mode the springs with spring constant $K_1$ vibrate while the springs with $K_2$ remain unstretched. In the second case it is the other way around.