In alpha particle scattering experiment, if \( v \) is the initial velocity of the particle, then the distance of closest approach is
- \( 4 \, d \)
- \( 2 \, d \)
- \( d \)
- \( \frac{d}{2} \)
The Correct Option is D
Solution and Explanation
In the alpha particle scattering experiment, the distance of closest approach is calculated using the relationship between the initial velocity of the particle and the distance \( d \) between the centers of the alpha particle and the nucleus:
\( r = \frac{d}{2} \)
This formula shows that the distance of closest approach is half the distance between the centers of the two particles when considering the initial velocity of the alpha particle.
Top Questions on Nuclear physics
- Write down the difference between nuclear fission and nuclear fusion.
- Bihar Board XII - 2025
- Physics
- 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
Questions Asked in KCET exam
- A particle is in uniform circular motion. The equation of its trajectory is given by \( x = 2t^2 - 3t + 5 \), where \( x \) and \( y \) are in meters. The speed of the particle is 2 m/s. When the particle attains the lowest \( y \)-coordinate, the acceleration of the particle is (in \( \text{m/s}^2 \)):
- KCET - 2025
- Friction
- If \( A \) is a square matrix of order \( 3 \times 3 \), \( \det A = 3 \), then the value of \( \det(3A^{-1}) \) is:
- KCET - 2025
- Matrices
- A potential at a point A is 3 V and that at another point B is 5 V. What is the work done in carrying a charge of 5 mC from B to A?
- KCET - 2025
- Electric Field
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
- KCET - 2025
- Newtons Laws of Motion
- A series LCR circuit containing an AC source of 100V has an inductor and a capacitor of reactances 24\(\Omega\) and 16\(\Omega\) respectively. If a resistance of 6\(\Omega\) is connected in series, then the potential difference across the series combination of inductor and capacitor will be
- KCET - 2025
- Electromagnetic waves