Problem set 33

1. The radius of the Fermi sphere of free electrons in a monovalent metal with an fcc structure, in which the volume of the unit cell is $a^3$, is
1. $\left(\frac{12\pi^2}{a^3}\right)^{1/3}$
2. $\left(\frac{3\pi^2}{a^3}\right)^{1/3}$
3. $\left(\frac{\pi^2}{a^3}\right)^{1/3}$
4. $\left(\frac{1}{a}\right)^{1/3}$
The radius of Fermi sphere is given by $k_f=\left(\frac{3\pi^2N}{V}\right)^{1/3}$. For FCC structure $\frac{N}{V}=\frac{4}{a^3}$. Hence, $k_f\left(\frac{12\pi^2}{a^3}\right)^{1/3}$. Hence, answer is (A).
2. Deviation from Rutherford scattering formula for $\alpha$-particle scattering gives an estimate of:
1. Size of an atom
2. Thickness of target
3. Size of a nucleus
4. half life of $\alpha$-emitter
Rutherford formula is derived by assuming nucleus to be point charge. However, nucleus is not a point charge. However, nucleus is not a point charge. There is a charge distribution inside the nucleus. Due to which there is change in angular dependence of scattering. From this information one can estimate size of nucleus.
Hence, answer is (C)
3. If the critical magnetic field for aluminium is $7.9\times 10^3 A/m$, the critical current which can flow through long thin superconducting wire of aluminium of diameter $1\times10^{-3}m$ is
1. 5.1 A
2. 1.6 A
3. 4.0 A
4. 2.48 A
Critical current is given by $$\begin{align*} I_c&=2\pi rH_c\\ &=2\times3.14*0.5\times10^{-3}\times7.9\times 10^3\\ &=2.48 A \end{align*}$$ Hence, answer is (D)
4. An elemental dielectric has a dielectric constant $\epsilon=12$ and it contains $5\times10^{18} atoms/m^3$. Its electronic polarizability in uints of $FM^2$ is
1. $4.17\times10^{-30}$
2. $8.34\times10^{-30}$
3. $6.32\times10^{-30}$
4. $12.64\times10^{-30}$
The Clausius Mossotti relation is $$\frac{\epsilon_r-1}{\epsilon_r+2}=\frac{N\alpha_e}{3\epsilon_0}$$ $$\begin{align*} \alpha_e&=\frac{\epsilon_r-1}{\epsilon_r+2}\frac{3\epsilon_0}{N}\\ &=\frac{12-1}{12+2}\frac{3\times8.85\times10^{-12Fm^{-1}}}{5\times10^{18} atoms/m^3}\\ &=4.17\times10^{-30}FM^2\\ \end{align*}$$ Hence, answer is (A)
5. Mangetite $(Fe_3O_4)$ has a cubic structure with a lattice constant of $8.4A^o$. The saturation magnetization in this material in units of A/m is :
1. $6.2\times10^5$
2. $6.2\times10^6$
3. $12.4\times10^6$
4. $12.4\times10^5$
As each $O$ has an oxidation number of $-2$ and there are 4 of them in the formula, the 3 Fe atoms must together have a charge of $+8$. This is consistent with their being two $Fe^{3+}$ ions and one $Fe^{2+}$ per formula unit. The ratio of $Fe^{3+}$ ions to $Fe^{2+}$ is 2 to 1. There are 8 $Fe_3O_4$ units per unit cell. Hence, there are 24 $Fe$ atoms in unit cell, out of which 16 are $Fe^{3+}$ and 8 are $Fe^{2+}$. Out of the 16 $Fe^{3+}$ ions 8 $Fe^{3+}$ ions are at octahedral position and 8 $Fe^{3+}$ ions are at tetrahedral position. The ions in tetrahedral sites oppose the applied magnetic field and the ions at octahedral positions reinforce the field. Consequently, $Fe^{3+}$ ions neutralizes the $Fe^{3+}$ at octahedral positions. In each $Fe^{3+}$ there are 5 unpaired $3d$ electrons and in each $Fe^{2+}$ there are 4 unpaired $3d$ electrons. Each unpaired electron has magnetic moment one Bohr magneton $\mu_B=9.27\times10^{-24}Am^2$. Hence, each $Fe^{2+}$ ion has 4$\mu_B$ magnetic moment. As, unit cell contains 8 $Fe^{2+}$ ions, total magnetic moment is $32\mu_B$.
Hence, saturation magnetization $$\begin{align*} M&=\frac{32\mu_B}{a^3}\\ &=\frac{32\times9.27\times10^{-24}}{\left(8.4\times10^{-10}\right)^3}\\ &=5.1\times10^5A/m \end{align*}$$