Question

The possible number of energy states in a Ge crystal containing $5\times10^{3}$ atoms is:

• A. $2\times10^{4}$
• B. $4\times10^{4}$
• C. $4\times10^{4}$
• D. $3\times10^{4}$
• E. $5\times10^{4}$

Hint:

For Group 14 elements (C, Si, Ge), the total states in the valence and conduction bands is $8N$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question relates to the band theory of solids, specifically for a semiconductor like Germanium (Ge). When individual atoms come together to form a crystal, their discrete atomic energy levels broaden into continuous energy bands due to interatomic interactions. We need to determine the number of available quantum states within these bands for a given number of atoms.
Step 2: Key Formula or Approach:
1. Germanium (Ge) is a Group 14 element, meaning each Ge atom has 4 valence electrons. These electrons occupy the valence band.
2. In a crystal containing N atoms, each atomic orbital contributes one energy state to the corresponding energy band.
3. The valence shell of Ge consists of s and p orbitals. One s-orbital and three p-orbitals give a total of 4 valence orbitals per atom.
4. Each orbital can hold 2 electrons (one spin-up, one spin-down), but band theory considers states. The valence band, formed from the s and p valence orbitals, will contain 4N available states. These states are completely filled by the 4N valence electrons at absolute zero temperature.
5. Similarly, the next higher band, the conduction band, also has a large number of available states (typically also considered 4N for Ge).
Step 3: Detailed Explanation:
The question asks for the "possible number of energy states". This is most commonly interpreted as the number of states in the valence band.
- Number of Ge atoms, \( N = 5 \times 10^3 \).
- Each Ge atom contributes its 4 valence orbitals (one 's' and three 'p') to form the valence band.
- Therefore, the total number of available energy states in the valence band is \( 4 \times N \).
- Number of states = \( 4 \times (5 \times 10^3) = 20 \times 10^3 = 2 \times 10^4 \).
Let's re-read the question. It asks for the "possible number of energy states". Maybe it means the total number of valence electrons? No, it asks for states. Let's reconsider the band formation. The s and p orbitals of N atoms form an s-band and a p-band. The s-band has 2N states and the p-band has 6N states. In semiconductors like Ge, these bands hybridize. The hybridized sp3 orbitals form a lower-energy valence band with 4N states and a higher-energy conduction band with 4N states. The valence band contains 4N states and is completely filled by the 4N valence electrons at 0 K. The conduction band contains 4N states and is completely empty at 0 K. So, the number of states in the valence band is \( 4 \times N = 4 \times (5 \times 10^3) = 2 \times 10^4 \). This matches option (A). Why is the given answer B (\( 4 \times 10^4 \))? Perhaps the question is asking for the total number of states in both the valence and conduction bands combined. Total states = States in valence band + States in conduction band = \( 4N + 4N = 8N \). Total states = \( 8 \times (5 \times 10^3) = 40 \times 10^3 = 4 \times 10^4 \). This calculation matches the answer key. This interpretation, while less common, is the only one that leads to the given answer. Step 4: Final Answer:
Assuming the question is asking for the total number of energy states in the valence and conduction bands combined, the calculation is as follows: - Number of atoms, \( N = 5 \times 10^3 \). - Number of states in valence band = \( 4N \). - Number of states in conduction band = \( 4N \). - Total states = \( 4N + 4N = 8N \). - Total states = \( 8 \times (5 \times 10^3) = 40 \times 10^3 = 4 \times 10^4 \).