Question:
The electron concentration \( n_e \) and hole concentration \( n_h \) in semiconductor are related to the number of intrinsic charge concentration \( n_i \) as
The electron concentration \( n_e \) and hole concentration \( n_h \) in semiconductor are related to the number of intrinsic charge concentration \( n_i \) as
Show Hint
In intrinsic semiconductors, the product of electron and hole concentrations is always equal to the square of the intrinsic charge concentration.
Updated On: Mar 6, 2025
- \( n_e n_h = n_i^2 \)
- \( n_e + n_h = n_i^2 \)
- \( n_e + n_h = 2n_i^2 \)
- \( n_e n_h = 2n_i \)
- \( n_e n_h^2 = n_i \)
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
In intrinsic semiconductors, the product of the electron concentration \( n_e \) and hole concentration \( n_h \) is equal to the square of the intrinsic carrier concentration \( n_i \): \[ n_e n_h = n_i^2 \] This relation holds because in an intrinsic semiconductor, the number of electrons is equal to the number of holes.
Thus, the correct answer is (A).
Was this answer helpful?
0
0
Top Questions on Nuclear physics
- Define the term, ‘distance of closest approach’. A proton of 3.95 MeV energy approaches a target nucleus \( Z = 79 \) in head-on position. Calculate its distance of closest approach.
- CBSE CLASS XII - 2025
- Physics
- Nuclear physics
Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \)
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given:- Mass of \( ^{235}_{92}\text{U} \): \( m(^{235}_{92}\text{U}) = 235.04393 \, \text{u} \)
- Mass of \( ^{140}_{54}\text{Xe} \): \( m(^{140}_{54}\text{Xe}) = 139.92164 \, \text{u} \)
- Mass of \( ^{94}_{38}\text{Sr} \): \( m(^{94}_{38}\text{Sr}) = 93.91536 \, \text{u} \)
- Mass of neutron \( ^{1}_0n \): \( m(^{1}_0n) = 1.00866 \, \text{u} \)
- Conversion factor: \( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)
- CBSE CLASS XII - 2025
- Physics
- Nuclear physics
- Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.
% Assertion Assertion (A): During the formation of a nucleus, the mass defect produced is the source of the binding energy of the nucleus.
Reason (R): For all nuclei, the value of binding energy per nucleon increases with mass number.
- CBSE CLASS XII - 2025
- Physics
- Nuclear physics
- Differentiate between ‘nuclear fission’ and ‘nuclear fusion’. Briefly discuss one example of each.
- CBSE CLASS XII - 2025
- Physics
- Nuclear physics
- An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.
- CBSE CLASS XII - 2025
- Physics
- Nuclear physics
View More Questions
Questions Asked in KEAM exam
- Benzene when treated with Br\(_2\) in the presence of FeBr\(_3\), gives 1,4-dibromobenzene and 1,2-dibromobenzene. Which type of reaction is this?
- KEAM - 2025
- Haloalkanes and Haloarenes
- 10 g of (90 percent pure) \( \text{CaCO}_3 \), treated with excess of HCl, gives what mass of \( \text{CO}_2 \)?
- KEAM - 2025
- Stoichiometry and Stoichiometric Calculations
- Evaluate the following limit: $ \lim_{x \to 0} \frac{1 + \cos(4x)}{\tan(x)} $
- KEAM - 2025
- Limits
- Given that \( \mathbf{a} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \mathbf{b} \), \( |\mathbf{a} + \mathbf{b}| = \sqrt{29} \), \( \mathbf{a} \cdot \mathbf{b} = ? \)
- KEAM - 2025
- Vector Algebra
- In an equilateral triangle with each side having resistance \( R \), what is the effective resistance between two sides?
- KEAM - 2025
- Combination of Resistors - Series and Parallel
View More Questions