Nuclear Physics

Mass Defect: Mass defect ( ) is the difference between the total mass of the individual nucleons (protons and neutrons) in a nucleus and the actual measured mass of the nucleus. It is given by: where:
is the number of protons,
is the number of neutrons,
and are the masses of a proton and a neutron, respectively,
is the actual nuclear mass.

Binding Energy: Binding energy ( ) is the energy required to break a nucleus into its constituent protons and neutrons. It is given by Einstein’s mass-energy equivalence: where: is the speed of light ( m/s),
is the mass defect.
Fission Process and Binding Energy Per Nucleon: Nuclear fission occurs when a heavy nucleus splits into two or more lighter nuclei, releasing a significant amount of energy. This is explained using the concept of binding energy per nucleon: For heavy nuclei (e.g., Uranium-235), the binding energy per nucleon is lower than that of medium-sized nuclei.
When a heavy nucleus splits, the resulting smaller nuclei have higher binding energy per nucleon, meaning energy is released in the process.
This released energy is the basis of nuclear power and atomic bombs.

The mass defect for a deuteron is the difference between the mass of the deuteron and the sum of the masses of its constituent nucleons (proton and neutron): Substitute the given values: The energy equivalent of the mass defect is: Thus, the mass defect is , and the energy equivalence is .

Mass Defect and Energy Released in the Fission of

In this fission process, the total mass before and after the reaction changes. The total mass defect is the difference between the mass of the fission products and the initial mass.

The initial mass is the mass of the nucleus plus the mass of the neutron:

Substituting the given values:

The final mass is the mass of the fission products (the and nuclei) plus the mass of the two neutrons:

Substituting the given values:

Now, the mass defect is:

To find the energy released, we use the equivalence :

Thus, the energy released in the process is .

Step 1: Calculate the Mass Defect Mass defect is given by: Substituting values:

Step 2: Calculate the Binding Energy Binding energy is given by: Substituting u: Thus, the mass defect is u, and the binding energy is MeV.

Distance of Closest Approach for a Proton Colliding with a Nucleus

The "distance of closest approach" refers to the minimum distance between the incident particle (in this case, a proton) and the nucleus due to electrostatic repulsion. This occurs when the kinetic energy of the proton is entirely converted into potential energy at the closest point.

We use the concept of energy conservation here:

The potential energy at the closest distance is given by the Coulomb force formula:

Where:

• is the atomic number of the nucleus (in this case, ),
• is the charge of the proton ( ),
• is the distance of closest approach,
• is the permittivity of free space ( ).

Now, equate the initial kinetic energy to the final potential energy:

The kinetic energy is given by:

Equating the two energies:

Substituting the known values and solving for :

After substituting values:

Solving for , we get:

Thus, the distance of closest approach is .

The curve of binding energy per nucleon as a function of mass number is a fundamental concept in nuclear physics. The binding energy per nucleon is a measure of the stability of a nucleus. As per nuclear physics, the binding energy per nucleon generally increases with the mass number until a particular peak, after which it starts to decrease.
The chart follows a specific trend:

• For lighter nuclei, the binding energy per nucleon increases rapidly with increasing mass number, as these smaller nuclei gain greater stability.
• This increase continues until we reach a peak at around mass numbers 56 to 58, corresponding to elements like iron and nickel. These elements are the most stable and have the highest binding energy per nucleon.
• After reaching this peak, the binding energy per nucleon gradually decreases for heavier nuclei. This is due to the increasing effect of the repelling force of protons at such high nucleon numbers, even though neutrons help to a certain extent to stabilize the nucleus.

Step 1: Expression for mass of the nucleus

The mass of a nucleus is approximately equal to the mass number times the mass of a nucleon (proton or neutron), i.e.,

Where:

• is the mass number (total number of protons and neutrons),
• is the mass of a single nucleon (approximately ).

Step 2: Expression for volume of the nucleus

The volume of a nucleus is related to its radius . The radius of the nucleus is given by the empirical formula:

Where is a constant approximately equal to .

The volume of a spherical nucleus is:

Substituting the expression for , we get:

Step 3: Expression for nuclear density

The nuclear density is defined as the mass per unit volume:

Substituting the expressions for and , we get:

Simplifying:

Step 4: Conclusion

Notice that in the final expression for , the mass number cancels out, and we are left with a constant value:

Therefore, the nuclear density is independent of the mass number. This shows that the density of the nucleus remains constant for all isotopes, regardless of their size or mass number.

Mass Defect: The mass defect of a nucleus is the difference between the total mass of its constituent nucleons (protons and neutrons) and the actual mass of the nucleus. The missing mass is converted into binding energy, which holds the nucleus together.

Binding Energy: The binding energy of a nucleus is the energy required to separate the nucleus into its individual nucleons. It is the energy equivalent of the mass defect.

Fission Process: In the fission process, a heavy nucleus splits into two smaller nuclei along with the release of energy. This occurs because the binding energy per nucleon of the smaller nuclei is greater than that of the original nucleus. The energy released in fission is due to the difference in binding energy per nucleon before and after the fission.

Thus, in fission, the total binding energy of the fragments after splitting is greater than the binding energy of the original nucleus, and this difference is released as energy.

Mass Defect, Binding Energy, and Fission Process

1. Mass Defect

The mass defect of a nucleus is the difference between the total mass of the individual nucleons (protons and neutrons) that make up the nucleus and the actual mass of the nucleus itself. Mathematically, it is expressed as:

2. Binding Energy

The binding energy of a nucleus is the energy required to separate the nucleus into its constituent protons and neutrons. It is related to the mass defect by Einstein’s equation:

Where is the mass defect, is the speed of light, and is the binding energy.

3. Fission Process

Nuclear fission is the process in which a heavy nucleus, such as uranium-235, splits into two or more smaller nuclei, releasing a large amount of energy. This occurs when a nucleus absorbs a neutron and becomes unstable. During fission:

• The nucleus splits into smaller fragments.
• A large amount of energy is released due to the increased binding energy per nucleon in the fission products.
• The binding energy per nucleon increases as the nucleus splits, resulting in energy release.

The circuit for studying the V-I characteristics of a p-n junction diode typically includes:
- A variable DC power supply to provide different values of voltage ,
- A p-n junction diode,
- A series resistor to limit the current through the diode, and
- A voltmeter and ammeter to measure the voltage across and the current through the diode, respectively.
Here is a simple diagram: The current and voltage are varied, and the corresponding V-I characteristics are plotted.

• For light nuclei (mass numbers less than 20), the binding energy per nucleon increases sharply.
• In the region of medium mass numbers, around the element iron, the curve reaches a maximum; this indicates that these nuclei are the most stable.
• For heavier nuclei beyond this point, the binding energy per nucleon gradually decreases because of increased repulsive forces among protons.

Number of Holes Created in P-Type Silicon Semiconductor Due to Doping

In a p-type silicon semiconductor, doping introduces **holes** (positive charge carriers) by substituting silicon atoms with dopant atoms. Let's calculate the number of holes created per cubic centimetre due to doping.

Given Data:

• The doping concentration is 1 dopant atom per silicon atoms.
• The number density of silicon atoms in the specimen is .

Step 1: Finding the Number of Dopant Atoms per Unit Volume

The number of dopant atoms per unit volume can be calculated by multiplying the doping concentration by the number density of silicon atoms:

Thus, the number of dopant atoms per unit volume is:

Step 2: Finding the Number of Holes per Cubic Metre

In a p-type semiconductor, each dopant atom introduces one hole. Therefore, the number of holes per cubic metre is equal to the number of dopant atoms per cubic metre:

Step 3: Converting to Cubic Centimetres

Since 1 cubic metre is equal to cubic centimetres, the number of holes per cubic centimetre is:

Conclusion:

The number of holes created per cubic centimetre in the p-type silicon semiconductor due to doping is .

Example of Dopants:

One common example of a dopant for creating p-type silicon is boron (B). Boron has one less valence electron than silicon, which results in the creation of a hole in the semiconductor.

Assertion and Reason Analysis for P-N Junction

Assertion (A):

We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of an n-type semiconductor.

Reason (R):

In a p-type semiconductor, while in an n-type semiconductor .

Explanation:

Let's break down the assertion and reason:

• In a p-type semiconductor, the majority charge carriers are holes ( ), and the minority carriers are electrons ( ), so .
• In an n-type semiconductor, the majority charge carriers are electrons ( ), and the minority carriers are holes ( ), so .
• When a p-type and n-type material are physically joined, a p-n junction is formed, but the assertion states that this junction cannot be formed by physically joining the materials, which is incorrect. A p-n junction is formed by the diffusion of electrons and holes at the interface between the two materials, creating a depletion region where recombination occurs.
• The reason ( in n-type and in p-type) is correct in terms of charge carrier concentrations in p-type and n-type semiconductors but does not explain the failure to form a p-n junction.

Conclusion:

Therefore, while both the assertion and the reason are true, the reason does not correctly explain the assertion. The assertion is false because a p-n junction can indeed be formed by physically joining a p-type and an n-type semiconductor.

Correct Answer:

Option 2: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Binding Energy per Nucleon vs. Mass Number

The binding energy per nucleon in a nucleus is a measure of how strongly the nucleons (protons and neutrons) are bound together. This energy is dependent on the mass number of the nucleus, and its trend as a function of shows an interesting pattern.

Trend of Binding Energy per Nucleon:

As a function of mass number , the binding energy per nucleon increases with up to iron ( ) and then decreases as increases further. This is because:

• Nuclei with mass number around (like iron) have the highest binding energy per nucleon, meaning they are the most stable nuclei.
• For lighter nuclei (with smaller ), the binding energy per nucleon increases as nucleons come together and form more stable configurations.
• For heavier nuclei (with larger ), the binding energy per nucleon decreases. This is because, as the nucleus becomes larger, the nuclear forces are less effective at binding the nucleons at the edges of the nucleus, and the overall energy required to hold the nucleus together becomes less efficient.

Conclusion:

The correct graph representing this behavior is one that shows a peak at (for iron) with the binding energy per nucleon increasing initially as increases and then decreasing for heavier nuclei.