Explanation:

In the given reaction, four protons combine to form the nucleus of a helium atom. This process is known as fusion, where smaller nuclei merge to create a larger one. During the annihilation process, two positrons and two electrons interact and transform into gamma rays. Therefore, the energy they stored due to their masses is released. The energy released can be calculated by using the following formula:

• E=mc^2

Since the mass of the positron and electron are the same:

Stars with a mass up to 4 times the mass of the sun will form a white dwarf. If their mass is between roughly 4 and 8 solar masses, they will form a neutron star. If the mass is greater than roughly 8 solar masses, they will form a black hole.

I: CORRECT Decreasing the wavelength is the same as increasing the frequency of the incident radiation. This increases the quantum of photon energy and results in the emission of electrons.

II: INCORRECT Higher intensity of the same light, has no effect on the electrons. Since each photon still carries the same energy as before.

III: CORRECT Reducing the threshold frequency might lead to a point where it is less than the frequency of the incident light (c/λ), and hence result in emission of electrons.

Explanation:

Effective management of nuclear waste requires preventing contamination of the soil, water supply, and atmosphere. Burying nuclear waste in the soil can lead to soil and water supply contamination. Similarly, disposing of it in the ocean poses numerous hazards to marine life. Burning nuclear waste can also result in contamination of the atmosphere.

The most suitable method for long-term management is storing the waste in a disused mine with appropriate containment measures.

Explanation:
A change in the proton composition of a nucleus is needed to label it as a different nucleus. Both alpha ($\alpha$) and Beta ($\beta$) decay change the proton and neutron number of the parent nucleus therefore a different daughter will be produced. On the other hand, gamma ($\gamma$) decay involves only the loss of excess energy, not the change in nucleon composition therefore the daughter is the same as the parent.

Therefore, the answer is C.

Explanation:

The star's position on the H-R diagram is determined by its temperature and luminosity. The star is hotter than the sun, it should fall to the left of the Sun’s position on the H-R diagram.

Furthermore, the star has a similar radius and hotter temperature compared to the Sun, resulting in higher luminosity according to the equation

• L=σAT^4

Consequently, it would fall above the Sun's position on the H-R diagram. By combining these two pieces of information, it can be deduced that the most probable position of the star is C.

Explanation:
Binding energy of $X$; (A)(E)=AE
By using E=mc^2, mass defect; m=AE/c^2
Therefore the answer is A.

Explanation:
Count all possible de-excitations below n=4.
From $n=4$ electron can de-excite to n=3,2,1 ; 3 possible frequencies.
From $n=3$ electron can de-excite to n=2,1; 2 possible frequencies.
From $n=2$ electron can de-excite to $n=1$; 1 possible frequency.

Total of 6 frequencies.

Explanation:
In the absence of the strong nuclear force there is no force to overcome repulsive electrostatic force between protons. Therefore, more protons in a large nucleus results in more repulsive electrostatic force.

Explanation:
The number of protons on both sides of the equation must be the same.

2+A=B+0

Therefore the answer is D.

Explanation:

Energy of the incident photon ($E=hf$), goes into:

1. releasing an electron (hfA and hfB respectively), and
2. imparting kinetic energy to the electron. Since stopping voltage converts this kinetic energy into electrical potential energy, this is measured by eVA and eVB for respective materials.

Since the second plate takes more energy to release the electron (∵fB>fA), it gets less kinetic energy after release, and hence smaller stopping potential.

Explanation:

I is incorrect: Photoelectric effect reflects the particle behavior of light.

II is correct: In the Davisson-Germer experiment electrons were scattered from atoms and these scattered electrons produced an interference pattern just like waves.

III is incorrect: Young's double slit experiment shows wave behavior of light

Explanation:

The decay constant depends on the radioactive material. It does not change for a given nuclide. In this case it is $\lambda$ for both samples as they are the same nuclide. Hence, I is incorrect.

Activity depends on the mass of the sample. If it is $A$ for mass $m$ at $t=0$, it will be $2A$ for $2m$. But after two half lives, it will be halved twice and become A/2. So, II is correct.

After one half life, half of $2m$ decays, and will leave a mass m decayed, while another m is active. After the second half life, half of the remaining $m$ decays leaving just another m/2 active. So the total mass of decayed material is 3m/2. Hence III is incorrect.

Explanation:

All photons carry the same energy (corresponding to the frequency of yellow light).

That means the maximum kinetic energy is the same for electrons in both cases. Hence equal stopping voltages.

Hence higher intensity implies larger number of photons per second, and thus larger number of electrons ejected, leading to a higher current. Thus i2>i1

Explanation:

A: From his postulates, Bohr showed that the energy of electrons is quantised. Hence, explained why they don't lose energy continuously. This explained why electrons don't spiral into the nucleus. So, A is correct.

B: Bohr's model works only for atoms and ions with single electrons. B is not true. Hence this is the correct choice.

C and D: Based on the assumption that angular momentum of atomic electrons is quantised, Bohr proved that their energies, and orbital sizes are also quantised. So, C and D are correct also.

Explanation:
I. is incorrect because control rods block neutrons and inhibit the rate of reaction

II. is incorrect because the moderator slows neutrons to increase the chance of being absorbed by a uranium nucleus and result in a fission.

III. is correct

Explanation:
$Q$ has a mass number of 234 and atomic number of 90. This means that Q has 234 nucleons, and 90 protons.

Since beta particles have mass number zero and charge number $-1$, the nucleus will gain a $+2$ charge while the mass number will remain unchanged. This gives a mass number of 234 and the charge number will become 92.

In the next step, the emission of 1 alpha particle will diminish the mass number by 4 and the charge number by 2 resulting in $R$ having a mass number of 230 and a charge number of 90.

Explanation:

With more kinetic energy, alpha particles can overcome the electrical potential of the nucleus, and get close enough to experience strong nuclear attraction that is responsible for the deviation from Rutherford's model.

A smaller proton number for the target nuclei means a lower positive charge, and hence a lower potential barrier to overcome. Again, this enables alpha particles to approach the nucleus much closer, and increases the chance of strong nuclear interaction, thus causing a deviation from Rutherford's prediction.

Explanation:
A represents the best conditions for this, in that hydrogen nuclei are smaller than helium nuclei.
B shows at least one reactant particle identical to the only product particle. This is not balanced.
C shows a larger particle splitting into smaller ones. This is called fission.
D is also unbalanced because 2 larger atoms would not combine to form a smaller one.

Explanation:
The moderator slows down neutrons so that they can cause successful fissions. Without a moderator, neutrons that are the product of other fissions have too much energy to interact with new nuclei such that they cause further fissions to occur.

Control rods block the neutrons to reduce the number of chain reaction fissions.

The heat exchanger transfers thermal energy to the external steam generator circuit.

A chain reaction is a process, not a specific material.

Explanation:

The stars have greater mass have higher luminosity which results in a shorter time to run out of hydrogen. Therefore, the lifetime of a star on the main sequence is roughly inversely proportional to the mass of the star. Since the star has greater mass than the sun, time is less than $T$.

Moreover, the mass of the star is less than four times the solar mass. Hence, it is a moderate-mass star and it will become a red giant in its next evolutionary stage.

Explanation:
A is a simple beta decay reaction.
B is an alpha decay reaction.
C is the type of fission reaction that occurs in a nuclear reactor for the generation of electricity with uranium as a fuel.
D shows a fission reaction that would require an input of energy to occur. It does not take place in a fission reactor.