Nuclear Physics

Radius of a nucleus:
This is a straight-line equation with slope .

Energy of one photon: Number of photons per second:

Step 1: Recall what is meant by nuclear fuel.
A nuclear fuel is a material that can undergo nuclear fission or fusion to release energy, typically used in nuclear reactors or nuclear weapons. The most common nuclear fuels are heavy isotopes that can sustain a chain reaction.
Step 2: Examine each option.
(1) Uranium:
Uranium (especially isotope U) is the most widely used nuclear fuel in both nuclear reactors and atomic bombs. It is capable of sustaining a nuclear fission chain reaction. (2) Thorium:
Thorium ( Th) is not fissile itself but is fertile, meaning it can be converted into fissile U in a reactor. Thorium-based reactors are being developed for future use. (3) Plutonium:
Plutonium (mainly Pu) is a man-made element used as a nuclear fuel in reactors and nuclear weapons. It is produced from uranium in reactors. (4) Titanium:
Titanium is a strong, lightweight metal known for its corrosion resistance and is used in aerospace, medical implants, and other industries. It does not undergo fission or fusion and is not used as a nuclear fuel.
Step 3: Conclusion Among the given options, uranium, thorium, and plutonium are all used as nuclear fuels, but titanium is not. Final Answer:

The distance of closest approach of an alpha particle moving towards a nucleus is given by: where is the atomic number of the nucleus, is the charge of the electron, and is the energy of the alpha particle. By substituting the known values and rearranging the equation, we find that the distance of closest approach is . Thus, the correct answer is .

The distance of closest approach occurs when the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy at the closest distance from the nucleus.
The initial kinetic energy of the alpha particle with linear momentum and mass is given by: $ between the alpha particle (charge ) and the nucleus (charge ) at a distance is given by: \) is Coulomb's constant.
At the distance of closest approach , the kinetic energy is equal to the potential energy: \) in terms of : \) is inversely proportional to the square of the linear momentum : \) and the distance of closest approach be .
Let the new linear momentum be and the new distance of closest approach be .
Using the proportionality: \) $ The distance of closest approach when the linear momentum is is .

The radius of a nucleus is given by the empirical formula: where is a constant with a value approximately m, and is the mass number of the nucleus. For , the radius is: First, compute : Therefore, Thus, the radius of the nucleus is m. Therefore, the correct answer is option (2), .

The nuclear reaction is valid as it satisfies the conservation of mass and atomic numbers.
Thus, the correct answer is Option 1.

Let the initial nucleus be . Emission of:

• One -particle decreases by 4 and by 2.
• Each -particle increases by 1.

For the final nucleus to be an \textit{isotope}, must remain unchanged. Let one and particles be emitted. Change in : So, one and two emissions restore , making the final nucleus an isotope (same , different ). Hence, option (3) is correct.

Step 1: Identify the given values.
Initial amount of radioactive material, g
Half-life of the material, years
Time for which the material is stored, years

Step 2: Use the formula for radioactive decay.
The amount of radioactive material remaining after time is given by: $ $
Therefore, the amount of remaining radioactive material after 4 years is 0.25 g.

Wolfgang Pauli predicted the existence of the neutrino to account for the missing energy and momentum in beta decay. Pauli suggested the existence of this particle in 1930, which was later confirmed experimentally.

The energy released in a nuclear reaction is given by Einstein’s mass-energy equivalence: If the speed of light becomes half, , then: So, energy decreases by:

The binding energy per nucleon is a measure of the stability of the nucleus. The higher the binding energy per nucleon, the more stable the nucleus is. has the highest binding energy per nucleon among all known elements. This makes it the most stable nucleus in terms of binding energy. Thus, the correct answer is .

The atom bomb works on the principle of uncontrolled nuclear fission. In this process, a heavy nucleus such as Uranium-235 or Plutonium-239 splits into two lighter nuclei, releasing a large amount of energy along with neutrons. These neutrons further trigger fission reactions, creating a chain reaction that releases massive explosive energy.

Step 1: Relationship Between Surface Area and Mass Number The surface area of a nucleus is proportional to the square of its radius: Since nuclear radius is related to mass number: Step 2: Compute the Mass Number Ratio Cubing both sides: Thus, the correct answer is .

The Rydberg formula for the wavelength of spectral lines of hydrogen is: where is the Rydberg constant, is the principal quantum number of the final state, and is the principal quantum number of the initial state ( ).
First Lyman line ( ): Transition from to .
So, .
Second Balmer line ( ): The Balmer series has .
First Balmer line is to .
Second Balmer line is to .
So, .
Ratio of wavelengths : The ratio .
This matches option (2).

Step 1: Energy-wavelength relation
Energy of photon and wavelength are related by: where , .
Step 2: Convert 5 eV to Joules
Step 3: Calculate wavelength
Step 4: Conclusion
The wavelength is approximately 248 nm.

Step 1: Understanding Bohr's Atomic Model Bohr's atomic model was developed to explain the structure of atoms and their energy levels. It successfully describes: - Quantized energy levels for electrons. - Emission spectra of hydrogen-like atoms (single-electron systems). Step 2: Applicability of Bohr’s Model Bohr’s model is valid only for hydrogenic atoms—atoms with a single electron, such as: - Hydrogen ( ), - Helium ion ( ), - Lithium ion ( ). For multi-electron atoms, electron-electron interactions make Bohr’s model inaccurate. Conclusion Thus, the correct answer is:

Step 1: Identify Given Information and Relevant Formulas
The mean life of a radioactive substance is given: minutes. The fraction of atoms disintegrated is given: . We need to find the time ( ) in minutes. The relationship between mean life ( ) and decay constant ( ) is: The radioactive decay law states that the number of undecayed nuclei ( ) at time ( ) is related to the initial number of nuclei ( ) by: The fraction of atoms disintegrated ( ) is given by: Step 2: Calculate the Decay Constant ( )
Using the given mean life: Step 3: Determine the Fraction of Atoms Remaining ( )
We are given the fraction of atoms disintegrated ( ). We can use this to find the fraction of atoms remaining: Rearrange the equation to solve for : Step 4: Equate and Solve for Time ( )
Now, use the decay law equation and the calculated fraction remaining: We know that can be written as . So: Comparing the exponents: Solve for : Substitute the value of from Step 2: Step 5: Analyze the Options
\begin{itemize} \item Option (1): 36. Incorrect. \item Option (2): 6. Incorrect. \item Option (3): 24. Correct, as it matches our calculated time. \item Option (4): 12. Incorrect. \end{itemize}

Step 1: Understanding Bohr's Atomic Model Bohr's atomic model was developed to explain the structure of atoms and their energy levels. It successfully describes: - Quantized energy levels for electrons. - Emission spectra of hydrogen-like atoms (single-electron systems). Step 2: Applicability of Bohr’s Model Bohr’s model is valid only for hydrogenic atoms—atoms with a single electron, such as: - Hydrogen ( ), - Helium ion ( ), - Lithium ion ( ). For multi-electron atoms, electron-electron interactions make Bohr’s model inaccurate. Conclusion Thus, the correct answer is:

The time period of the revolution of an electron in an orbit is proportional to the cube of the principal quantum number, . Therefore, the ratio of time periods for the second and third excited states is: $ $

The Rydberg formula for the wavelength of spectral lines of hydrogen is: where is the Rydberg constant, is the principal quantum number of the final state, and is the principal quantum number of the initial state ( ).
First Lyman line ( ): Transition from to .
So, .
Second Balmer line ( ): The Balmer series has .
First Balmer line is to .
Second Balmer line is to .
So, .
Ratio of wavelengths : The ratio .
This matches option (2).

The frequency of the Lyman series transition from level to 1 is given by the Rydberg formula:
- For the first Lyman line ( ):
- For the second Lyman line ( ):
The difference between the frequencies is:
Hence, the difference is .

Step 1: Known Information.
Nuclear matter refers to the material inside atomic nuclei, which is extremely dense.
The density of nuclear matter is a well-known physical quantity and is typically on the order of .
Step 2: Understanding Nuclear Density.
Nuclear density is derived from the fact that all nuclei have approximately the same density, regardless of their size. This is because the volume of a nucleus scales with the number of nucleons (protons and neutrons), and the mass also scales linearly with the number of nucleons. Thus, the density remains constant across different nuclei. The approximate value of nuclear density is: $ $ Final Answer: