LEVELJEE Main
Visualized Solution
The Sigma Insight: Nucleus and Nuclear Reaction
Unlocking the Secrets of the Nucleus
Imagine you have a perfectly built Lego castle. Now, imagine you take it apart and weigh all the individual Lego bricks. You would expect the total weight of the individual bricks to be exactly the same as the weight of the fully assembled castle, right? In our everyday macroscopic world, this is absolutely true. But when we dive into the quantum realm of the atomic nucleus, this intuitive logic completely shatters.
Let's explore this mind-bending phenomenon through the lens of an Oxygen-17 isotope, denoted as .
Analyzing the Setup
First, we need to understand what exactly is inside this specific nucleus. The atomic number () of Oxygen is . This tells us that there are exactly protons residing in the nucleus. The mass number () is , which represents the total number of nucleons (protons plus neutrons).
To find the number of neutrons (), we simply subtract the number of protons from the total mass number:
So, our Oxygen-17 nucleus is a tightly packed cluster of protons and neutrons.
The Mystery of the Missing Mass
Now, let's do a thought experiment. Suppose we could place our intact Oxygen-17 nucleus on a highly precise quantum scale. Let's call its mass .
Next, we take the nucleus apart and weigh the protons and neutrons individually. The mass of a single proton is , so protons weigh . The mass of a single neutron is , so neutrons weigh . The total mass of these separated, free nucleons is:
Here is where the universe plays a trick on us: the mass of the intact nucleus () is actually less than the sum of the masses of its individual parts (). This difference is known as the mass defect ().
The Master Equation
Binding Energy
Where did this missing mass go? It didn't just vanish into thin air. According to Albert Einstein's legendary mass-energy equivalence principle, mass and energy are two sides of the same coin, related by the equation .
When the protons and neutrons come together to form the nucleus, the strong nuclear force binds them tightly. In doing so, a small amount of their mass is converted into a massive amount of energy, which is released into the universe. This is the Binding Energy.
To calculate this energy, we multiply the mass defect by the speed of light squared (). In physics, binding energy is often expressed as the energy of the bound state minus the energy of the free state, which gives a negative value indicating a stable, lower-energy system.
Therefore, the expression for the nuclear binding energy is:
Expanding the negative sign, we get our final, elegant expression:
This formula beautifully encapsulates the profound truth that the whole is sometimes literally less than the sum of its parts, bound together by the fundamental forces of the cosmos.
Similar Questions
JEE Advanced 2016
LEVELJEE Advanced
The electrostatic energy of protons uniformly distributed throughout a spherical nucleus of radius is given by The measured masses of the neutron, , and are , , and , respectively. Given that the radii of both the and nuclei are same, ( is the speed of light) and . Assuming that the difference between the binding energies of and is purely due to the electrostatic energy, the radius of either of the nuclei is ()
(A)
2.85 fm
(B)
3.03 fm
(C)
3.42 fm
(D)
3.80 fm
JEE Advanced 2022
LEVELJEE Advanced
The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be and the binding energy of a neutron be in the nucleus. Which of the following statement(s) is(are) correct?
* Multiple Correct Options
(A)
is proportional to where is the atomic number of the nucleus.
(B)
is proportional to where is the mass number of the nucleus.
(C)
is positive.
(D)
increases if the nucleus undergoes a beta decay emitting a positron.
LEVELJEE Main
If the binding energy per nucleon in and nuclei are and respectively, then in the reaction energy of proton must be
(A)
(B)
(C)
(D)
JEE Main 2020
LEVELJEE Main
Find the binding energy per nucleon for . Mass of proton , mass of neutron and mass of tin nucleus . (Take, )
(A)
9.0 MeV
(B)
7.5 MeV
(C)
8.0 MeV
(D)
8.5 MeV
LEVELJEE Advanced
Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron kg, mass of proton kg, mass of electron kg)
(A)
MeV
(B)
MeV
(C)
MeV
(D)
MeV
JEE Main 2020
LEVELJEE Main
The radius of a nucleus of mass number can be estimated by the formula m. It follows that the mass density of a nucleus is of the order of ( kg)
(A)
(B)
(C)
(D)
JEE Advanced 2007
LEVELJEE Main
In the options given below, let denote the rest mass energy of a nucleus and a neutron. The correct option is
(A)
(B)
(C)
(D)
LEVELJEE Advanced
Let be the mass of proton, the mass of neutron. the mass of nucleus and the mass of nucleus. Then
* Multiple Correct Options
(A)
(B)
(C)
(D)
LEVELJEE Main
Comprehension Passage
A nucleus of mass is at rest and decays into two daughter nuclei of equal mass each. Speed of light is .
Question 1:
The binding energy per nucleon for the parent nucleus is and that for the daughter nuclei is . Then,
(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main
