Explanation​:

To be a stable star, outwards and inwards forces must be balanced. Outwards force is the result of the pressure of radiation and temperature while the inwards force is formed due to gravitational attraction.

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​:
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​:
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.

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​:

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

2+A=B+0

Therefore the answer is D.

Explanation​:

The SI unit of radioactive decay constant is s^−1, which is the same as that of frequency.

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.

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.

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​:

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:

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​:

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.

In beta decays, the continuous energy spectra and apparent violation of the law of conservation of momentum let to the postulation of the existence of the neutrino as a particle that could account for the anomaly. Therefore the answer is C.

Nuclear energy levels are evidenced by the discrete energies of alpha particle emissions along with the discrete energies of gamma emissions caused by the nucleus transitioning to lower energy states.

Absorption and emission spectra give evidence of atomic energy levels.

Positrons were first observed in cloud chamber experiments involving cosmic rays.

The wave model assumes that the energy from the light is delivered in one continuous wave. As the wavelengths of the lasers are equal, the property that would increase intensity would be the amplitude of the light wave.

In the particle model, light is composed of photons. The photon energy for each laser is equal, so a higher intensity is achieved by when there are more photons emitted per second.

Explanation​:
A.  The drop in the count rate is considerable but does not completely drop off to zero. The remaining count can easily be explained by the presence of background radiation. This indicates that the emission from the source is almost completely blocked by the sheet of paper. Alpha is the only type of radiation that can be nearly completely stopped by a sheet of paper.

B.  If both alpha and beta were being emitted in equal numbers, the sheet of paper would block the alpha but an appreciable amount would penetrate through the paper as beta radiation. So the count rate after the paper is inserted should be larger than the recorded value.

C.  If only beta radiation was emitted, the count rate would not have significantly diminished after the paper was inserted because beta radiation is able to penetrate through paper and if that is the only type of radiation, there is no reason for the count to fall that far.

D.  The information is sufficient to draw a reasonable conclusion. That conclusion is expressed in answer A.

Explanation​:

The aluminum nucleus has fewer protons than the gold nucleus hence the effects of repulsion are less for the aluminum nucleus when compared with the gold.

The particles will be able to have a closer approach to the aluminum, and the gold will cause more scattering than the aluminum due to its stronger charge.