In a hypothetical fission reaction
\(^{92}X^{236} \rightarrow ^{56}Y^{141} + ^{36}Z^{92} + 3R\) The identity of emitted particles (R) is:
\(^{92}X^{236} \rightarrow ^{56}Y^{141} + ^{36}Z^{92} + 3R\) The identity of emitted particles (R) is:
- Proton
- Electron
- Neutron
- \(\gamma\)-radiations
The Correct Option is C
Solution and Explanation
To identify the emitted particles, let us verify the conservation of atomic number (\(Z\)) and mass number (\(A\)).
- Atomic number (\(Z\)):
\[ Z_{\text{LHS}} = 92, \quad Z_{\text{RHS}} = 56 + 36 = 92 \]
\(Z\) is conserved.
- Mass number (\(A\)):
\[ A_{\text{LHS}} = 236, \quad A_{\text{RHS}} = 141 + 92 = 233 \]
The mass number is not conserved. The difference is:
\[ A_{\text{LHS}} - A_{\text{RHS}} = 236 - 233 = 3 \]
The missing mass corresponds to three neutrons (\(R = \text{neutrons}\)).
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
Questions Asked in JEE Main exam
- Let \( f(x) = \frac{x - 5}{x^2 - 3x + 2} \). If the range of \( f(x) \) is \( (-\infty, \alpha) \cup (\beta, \infty) \), then \( \alpha^2 + \beta^2 \) equals to.
- JEE Main - 2025
- Functions
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics