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
- An atom \( ^8_3X \) is bombarded by a shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleus is recorded by:
- JEE Main - 2026
- Physics
- Nuclear physics
A small bob A of mass m is attached to a massless rigid rod of length 1 m pivoted at point P and kept at an angle of 60° with vertical. At 1 m below P, bob B is kept on a smooth surface. If bob B just manages to complete the circular path of radius R after being hit elastically by A, then radius R is_______ m :
- JEE Main - 2026
- Physics
- Nuclear physics
- The minimum frequency of photon required to break a particle of mass $15.348$ amu into $4$ particles is __________ kHz.
[Mass of He nucleus $=4.002$ amu, $1$ amu $=1.66\times10^{-27}$ kg, $h=6.6\times10^{-34}$ J$\cdot$s and $c=3\times10^8$ m/s]- JEE Main - 2026
- Physics
- Nuclear physics
- A \(7.9\ \text{MeV}\) \(\alpha\)-particle scatters from a target material of atomic number \(79\). From the given data, the estimated diameter of the nuclei of the target material is (approximately) _________ m. \[ \left[\frac{1}{4\pi\varepsilon_0}=9\times10^9\ \text{Nm}^2\text{/C}^2 \text{ and electron charge }=1.6\times10^{-19}\ \text{C}\right] \]
- JEE Main - 2026
- Physics
- Nuclear physics
- Mass number of a nucleus is \(\alpha\) and its radius is \(R_\alpha\). Radius of another nucleus of mass number \(\beta\) is \(R_\beta\). If \(\beta = 8\alpha\), then find \(\dfrac{R_\alpha}{R_\beta}\).
- JEE Main - 2026
- Physics
- Nuclear physics
Questions Asked in JEE Main exam
- The system of linear equations
$x + y + z = 6$
$2x + 5y + az = 36$
$x + 2y + 3z = b$
has- JEE Main - 2026
- Matrices and Determinants
Method used for separation of mixture of products (B and C) obtained in the following reaction is:
- JEE Main - 2026
- p -Block Elements
- The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (\(R \le L\)) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M) :
- JEE Main - 2026
- Kinematics
Which of the following best represents the temperature versus heat supplied graph for water, in the range of \(-20^\circ\text{C}\) to \(120^\circ\text{C}\)?
- JEE Main - 2026
- Thermodynamics
- When a part of a straight capillary tube is placed vertically in a liquid, the liquid rises upto certain height \( h \). If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by \(1%\) each, then the height of the liquid in the tube will change by ________ %.
- JEE Main - 2026
- thermal properties of matter
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.