Discussion
The kinetic energy of a free electron at a corner of the first Brillouin zone of a two-dimensional square lattice is larger than that of an electron at the midpoint of a side of the zone by a factor b. The value of b is :
Answer: B
The factor is b = 2.
For a square lattice of side a, the first Brillouin zone is a square with corners at \((\pm\pi/a, \pm\pi/a)\) and side midpoints at \((\pm\pi/a, 0)\) and \((0, \pm\pi/a)\). A free electron's kinetic energy is \(\hbar^{2}k^{2}/2m\), so
\[ \begin{aligned} k_{corner}^{2} &= 2\left(\frac{\pi}{a}\right)^{2} \\ k_{mid}^{2} &= \left(\frac{\pi}{a}\right)^{2} \\ b &= \frac{k_{corner}^{2}}{k_{mid}^{2}} = 2 \end{aligned} \]
\(\sqrt{2}\) is the ratio of the wave vectors, not of the energies; 4 and 8 square that ratio once or twice too often.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development