Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 105 and 115. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be :
Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 105 and 115. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be :
- 0.8 MeV
- 275 MeV
- 220 MeV
- 176 MeV
The Correct Option is D
Solution and Explanation
220A → 105B+115C
⇒ Q = [105 × 6.4 + 115×6.4] – [220 × 5.6] MeV
⇒ Q = 176 MeV
The correct option is (D) : 176 MeV
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
- In a group of 3 girls and 4 boys, there are two boys \( B_1 \) and \( B_2 \). The number of ways in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but \( B_1 \) and \( B_2 \) are not adjacent to each other, is:
- JEE Main - 2025
- Permutations
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
- A galvanometer having a coil of resistance 30 \( \Omega \) needs 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be \( X \, \Omega \), where \( X \) is:
- JEE Main - 2025
- Electrical Circuits and Shunt Resistance
- The variance of the numbers 8, 21, 34, 47, \dots, 320, is:
- JEE Main - 2025
- Arithmetic Progression and Variance
- Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : If Young’s double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer.
Reason (R) : The speed of light reduces in an optically denser medium than air while its frequency does not change.
In the light of the above statements, choose the most appropriate answer from the options given below :
Concepts Used:
Nuclear Physics
Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions, in addition to the study of other forms of nuclear matter. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons
Radius of Nucleus
‘R’ represents the radius of the nucleus. R = RoA1/3
Where,
- Ro is the proportionality constant
- A is the mass number of the element
Total Number of Protons and Neutrons in a Nucleus
The mass number (A), also known as the nucleon number, is the total number of neutrons and protons in a nucleus.
A = Z + N
Where, N is the neutron number, A is the mass number, Z is the proton number
Mass Defect
Mass defect is the difference between the sum of masses of the nucleons (neutrons + protons) constituting a nucleus and the rest mass of the nucleus and is given as:
Δm = Zmp + (A - Z) mn - M
Where Z = atomic number, A = mass number, mp = mass of 1 proton, mn = mass of 1 neutron and M = mass of nucleus.