The Bloch theorem states that within a crystal, the wave function $$\psi \left( {\overrightarrow {\bf{r}} } \right)$$ , of an electron has the form
A. $$\psi \left( {\overrightarrow {\bf{r}} } \right) = u\left( {\overrightarrow {\bf{r}} } \right){e^{i.\overrightarrow {\bf{k}} .\overrightarrow {\bf{r}} }}$$ where, $$u\left( {\overrightarrow {\bf{r}} } \right)$$ is an arbitrary function and $$\overrightarrow {\bf{k}} $$ is an arbitrary vector
B. $$\psi \left( {\overrightarrow {\bf{r}} } \right) = u\left( {\overrightarrow {\bf{r}} } \right){e^{i.\overrightarrow {\bf{G}} .\overrightarrow {\bf{r}} }}$$ where, $$u\left( {\overrightarrow {\bf{r}} } \right)$$ is an arbitrary function and $$\overrightarrow {\bf{G}} $$ is a reciprocal lattice vector
C. $$\psi \left( {\overrightarrow {\bf{r}} } \right) = u\left( {\overrightarrow {\bf{r}} } \right){e^{i.\overrightarrow {\bf{G}} .\overrightarrow {\bf{r}} }}$$ where $$u\left( {\overrightarrow {\bf{r}} } \right) = u\left( {\overrightarrow {\bf{r}} + \overrightarrow {\bf{A}} } \right),\,\overrightarrow {\bf{A}} $$ is a lattice and $$\overrightarrow {\bf{G}} $$ is a reciprocal lattice vector
D. $$\psi \left( {\overrightarrow {\bf{r}} } \right) = u\left( {\overrightarrow {\bf{r}} } \right){e^{i.\overrightarrow {\bf{k}} .\overrightarrow {\bf{r}} }}$$ where, $$u\left( {\overrightarrow {\bf{r}} } \right) = u\left( {\overrightarrow {\bf{r}} + \overrightarrow {\bf{A}} } \right),\,\overrightarrow {\bf{A}} $$ is a lattice vector and $$\overrightarrow {\bf{k}} $$ is an arbitrary vector
Answer: Option D
The valence electrons do not directly determine the following property of a metal
A. electrical conductivity
B. thermal conductivity
C. shear modulus
D. metallic lustre
A. $${\left( {\frac{{2Q}}{P}} \right)^{ - 6}}$$
B. $${\left( {\frac{Q}{P}} \right)^{ - 6}}$$
C. $${\left( {\frac{P}{{2Q}}} \right)^{ - 6}}$$
D. $${\left( {\frac{P}{Q}} \right)^{ - 6}}$$
A. $$N\mu \coth \left( {\frac{{\mu B}}{{{k_B}T}}} \right)$$
B. $$N\mu \tanh \left( {\frac{{\mu B}}{{{k_B}T}}} \right)$$
C. $$N\mu \sinh \left( {\frac{{\mu B}}{{{k_B}T}}} \right)$$
D. $$N\mu \cosh \left( {\frac{{\mu B}}{{{k_B}T}}} \right)$$
A. $$\sqrt {2C\left( {\frac{1}{{{M_1}}} + \frac{1}{{{M_2}}}} \right)} $$
B. $$\sqrt {C\left( {\frac{1}{{2{M_1}}} + \frac{1}{{{M_2}}}} \right)} $$
C. $$\sqrt {C\left( {\frac{1}{{{M_1}}} + \frac{1}{{2{M_2}}}} \right)} $$
D. zero
Join The Discussion