Problem set 31

1. The two dimensional lattice of graphene is an arrangement of Carbon atoms forming a honeycomb lattice of lattice spacing $a$, as shown below. The Carbon atoms occupy the vertices.
   
1. The Wigner-Seitz cell has an area of
1. $2a^2$
2. $\frac{\sqrt{3}}{2}a^2$
3. $6\sqrt{3}a^2$
4. $\frac{3\sqrt{3}}{2}a^2$
2. The Bravais lattice for this array is a
1. rectangular lattice with basis vectors $\vec d_1$ and $\vec d_2$
2. rectangular lattice with basis vectors $\vec c_1$ and $\vec c_2$
3. hexagonal lattice with basis vectors $\vec a_1$ and $\vec a_2$
4. hexagonal lattice with basis vectors $\vec b_1$ and $\vec b_2$
Wigner-Seitz cell contains one lattice point. Hence, volume of Wigner-Seitz cell is equal to volume of any normal cell divided by the number of lattice points in the normal cell ($N$). $${\scriptstyle\text{Area of Wigner-Seitz cell }=\frac{\text{area of normal cell}}{N}}$$ Here, normal cell is hexagonal, its area is $3\sqrt{3}a^2$ and number of lattice points $N=6\times\frac{1}{3}=2$. $$\text{Area of Wigner-Seitz cell }=\frac{3\sqrt{3}a^2}{2}$$
Hence, answer is (D)
The Bravais lattice for this array is a hexagonal lattice with basis vectors $\vec b_1$ and $\vec b_2$.
Hence, answer is (D)
2. A narrow beam of X-rays with wavelength $1.5 A^\circ$ is reflected from an ionic crystal with an fcc lattice structure with a density of $3.32 gm cm^{-3}$. The molecular weight is $108 AMU$ (1 AMU $= 1.66 \times10^{-24} g$).
1. The lattice constant is
1. $6.00A^\circ$
2. $4.56A^\circ$
3. $4.00A^\circ$
4. $2.56A^\circ$
2. The sine of the angle corresponding to $(111)$
1. $\sqrt{3}/4$
2. $\sqrt{3}/8$
3. $1/4$
4. $1/8$
For FCC structure number of lattice points per unit cell $n=4$ $$\rho=3.32 gm cm^{-3}=3320kg/m^3$$ $$M=108 AMU$$ Now, molecular weight is the weight of the 1 mole of molecules (1 mole of molecules means $N_0=6.02322\times10^{23}$ number of molecules i.e. Avogadro's number).
As, $N_0$ molecules have molecular weight $M$, $n$ molecules in the unit cell will have weight $\frac{n\times M}{N_0}$
$${\scriptstyle\text{Density of unit cell}=\frac{\text{weight of unit cell}}{\text{volume of unit cell}}}$$ $$\rho=\frac{n\times M}{N_0a^3}$$ $$\begin{align*} a^3&=\frac{n\times M}{N_0\rho}\\ &=\frac{4\times 108}{6.02322\times 10^{23}\times 3320}\\ &=6A^o \end{align*}$$ Hence, answer is (A)
According to Bragg's law $$2d\sin\theta=\lambda\Rightarrow\sin\theta=\frac{\lambda}{2d}$$ $$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ For (111) plane $d=\frac{a}{\sqrt{3}}$ $$\sin\theta=\frac{\lambda\sqrt{3}}{2a}=\frac{1.5\times\sqrt{3}}{2\times6}=\frac{\sqrt{3}}{8}$$ Hence, answer is (B)
3. If an electron is in the ground state of the hydrogen atom, the probability that its distance from the proton is more than one Bohr radius is approximately
1. 0.68
2. 0.48
3. 0.28
4. 0.91
Ground state wave function is $\psi=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$. Probability that electron is found within Bohr radius is given by $$\begin{align*} &P(r)=\int\limits_0^{a_0}|\psi|^2\:4\pi r^2 dr\\ & =\frac{4}{a_0^3}\int\limits_0^{a_0}r^2e^{-2r/a_0}\:dr\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{\left[r^2\int e^{-2r/a_0}\:dr\right]_0^{a_0}-\left[\int 2r \left(\int e^{-2r/a_0}\:dr\right)\:dr\right]_0^{a_0}\right\}}\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{\left[r^2e^{-2r/a_0}\left(-\frac{a_0}{2}\right)\right]_0^{a_0}+a_0\left[\int r e^{-2r/a_0}\:dr\right]_0^{a_0}\right\}}\\ &{\scriptscriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}+a_0\left[r\int e^{-2r/a_0}\:dr-\int e^{-2r/a_0}\left(-\frac{a_0}{2}\right)\:dr\right]_0^{a_0}\right\}}\\ &{\scriptscriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}+a_0\left[re^{-2r/a_0}\left(-\frac{a_0}{2}\right)-e^{-2r/a_0}\left(\frac{a_0^2}{4}\right)\right]_0^{a_0}\right\}}\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}-\frac{a_0^3e^{-2}}{2}-\frac{a_0^3e^{-2}}{4}+\frac{a_0^3}{4}\right\}}\\ &=-\left\{-5\times\frac{1}{e^2}+1\right\}=0.32 \end{align*}$$ Hence, probability of finding electron outside Bohr radius=1-0.32=0.68
Hence, answer is (A)