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Animated Solution for Physics - Semiconductors: If the lattice constant of this semiconductor is decreased, then which of the following is correct?

Select Answer:

Visualized Solution

Understanding the Energy Band Diagram

The diagram illustrates the energy bands of a semiconductor.
$E_{c}$ represents the width of the conduction band.
$E_{v}$ represents the width of the valence band.
$E_{g}$ represents the forbidden energy gap, or band gap.

Effect of Lattice Constant

The lattice constant is the average distance between adjacent atoms in the crystal lattice.
Decreasing the lattice constant means the atoms are being compressed and brought closer together.

Variation of Band Gap ( $E_{g}$ )

As the interatomic distance decreases below the equilibrium lattice constant, the strong repulsion between inner electron shells increases.
This intense interaction causes the energy levels to shift further apart, thereby increasing the band gap $E_{g}$ .

Variation of Band Widths ( $E_{c}$ and $E_{v}$ )

Concurrently, as the band gap widens due to this specific shifting of energy levels, the spread of the allowed energy states becomes restricted.
Consequently, the widths of the allowed bands, $E_{c}$ and $E_{v}$ , decrease.

Conclusion

Therefore, $E_{c}$ and $E_{v}$ decrease, but $E_{g}$ increases.
This matches option (c).

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The Sigma Insight: Energy Bands in Solids and Semiconductors

Solution Diagram

Analyzing the Setup

To truly appreciate this problem, we first need to understand the anatomy of the energy band diagram presented to us. In a solid semiconductor, electrons are not free to take on just any energy value. Instead, they are confined to specific ranges of energy called allowed bands.

At the bottom of our diagram, we have the valence band, which is typically full of electrons. The width of this band is denoted by

E_{v}

. Above it lies the conduction band, which is mostly empty at absolute zero but can accept excited electrons. Its width is denoted by

E_{c}

. Separating these two allowed regions is a strict no-go zone for electrons, known as the forbidden energy gap or band gap, labeled as

E_{g}

.

The Physics of Lattice Compression

The question asks us to imagine a scenario where the lattice constant of the semiconductor is decreased. But what exactly does that mean? The lattice constant is simply the average physical distance between adjacent atoms in the crystal structure. Decreasing it is akin to applying immense pressure to the crystal, forcing the atoms to pack closer together than they naturally would.

Imagine a crowded room where everyone is standing at a comfortable distance. This comfortable distance is the equilibrium lattice constant. If the walls of the room start closing in, people are forced closer together. Initially, this might just lead to more interaction (overlapping of electron clouds). However, as they are pushed uncomfortably close, strong repulsive forces take over.

In the quantum world of the semiconductor, when atoms are compressed beyond their equilibrium distance, the intense repulsion between their inner electron shells drastically alters the potential energy landscape of the crystal. This strong interaction forces the energy levels to shift and split further apart. As a direct consequence, the separation between the highest valence energy state and the lowest conduction energy state increases. In other words, the band gap $E_{g}$ increases.

The Final Verdict

But what happens to the widths of the allowed bands themselves? As the band gap widens due to this extreme compression and shifting of energy levels, the spread of the allowed energy states becomes more restricted. The energy bands effectively narrow down.

Therefore, the width of the conduction band (

E_{c}

) and the width of the valence band (

E_{v}

) both decrease.

Summarizing our physical thought experiment: compressing the lattice (decreasing the lattice constant) causes the band gap

E_{g}

to increase, while simultaneously causing the band widths

E_{c}

and

E_{v}

to decrease. This elegant interplay of quantum mechanics and solid-state physics leads us directly to the correct conclusion, which perfectly matches option (c).

Common Traps & FAQs

Why does the band gap increase when atoms get closer?

When atoms are compressed beyond their equilibrium lattice constant, the strong repulsion between their inner electron shells causes the energy levels to shift apart, widening the forbidden gap between the valence and conduction bands.

Shouldn't the band widths increase because of more overlap?

While initial overlap at larger distances increases band width, compressing the lattice beyond the equilibrium point alters the potential landscape significantly, which can lead to a narrowing of the allowed energy bands as the gap widens.

What exactly are

E_{c}

and

E_{v}

in this diagram?

In this specific diagram,

E_{c}

and

E_{v}

represent the energy widths of the conduction and valence bands, respectively, not just the energy levels of their edges.

Is this behavior true for all materials?

This specific behavior (increasing

E_{g}

and decreasing band widths) is characteristic of certain semiconductors under compression, depending on their specific band structure and the nature of their chemical bonds.

How does this relate to the equilibrium lattice constant?

The equilibrium lattice constant is where the total energy of the crystal is minimized. Changing the lattice constant by applying pressure moves the system away from this minimum, drastically altering the energy band structure.

Similar Questions

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 nm$ , is incident on it. The band gap in (eV) for the semiconductor is

1.1 eV

2.5 eV

0.5 eV

0.7 eV

For extrinsic semiconductors when doping level is increased,

Fermi level of p-type semiconductor will go upward and Fermi level of n-type semiconductors will go downward

Fermi level of p-type semiconductors will go downward and Fermi level of n-type semiconductor will go upward

Fermi level of p and n-type semiconductors will not be affected

Fermi level of both p-type and n-type semiconductors will go upward for

T > T_{F}

K and downward for

T < T_{F}

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is Fermi temperature

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0.9

0.7

0.5

1.1

Statement I: By doping silicon semiconductor with pentavalent material, the electrons density increases. Statement II: The n-type semiconductor has net negative charge. In the light of the above statements, choose the most appropriate answer from the options given below.

Statement I is true but statement II is false.

Statement I is false but statement II is true.

Both statement I and statement II are true.

Both statement I and statement II are false.

Which of the following statements is not true ?

The resistance of intrinsic semiconductors decreases with increase of temperature.

Doping pure Si with trivalent impurities give

p

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The majority carriers in

n

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p

n

junction can act as a semiconductor diode.

In a semiconductor, the number density of intrinsic charge carriers at $2 7^{\circ} C$ is $1.5 \times 1 0^{16} / m^{3}$ . If the semiconductor is doped with impurity atom, the hole density increases to $4.5 \times 1 0^{22} / m^{3}$ . The electron density in the doped semiconductor is ...... $\times 1 0^{9} / m^{3}$ .

The energy band gap is maximum in

superconductors

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an $n$ -type semiconductor, the density of electrons is $1 0^{19} m^{- 3}$ and their mobility is $1.6 m^{2} (V-s)^{- 1}$ , then the resistivity of the semiconductor (since, it is an $n$ -type semiconductor contribution of holes is ignored) is close to

2 Ω -m

0.2 Ω -m

0.4 Ω -m

4 Ω -m

If the ratio of the concentration of electrons to that of holes in a semiconductor is $\frac{7}{5}$ and the ratio of currents is $\frac{7}{4}$ , then what is the ratio of their drift velocities?

\frac{5}{8}

\frac{4}{5}

\frac{5}{4}

\frac{4}{7}

A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of

each of them increases

each of them decreases

copper decreases and germanium increases

copper increases and germanium decreases