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

2+A=B+0

Therefore the answer is D.

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​:

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​:

By considering the position of the red giants and the Sun on an H-R diagram, it is easy to say that red giants have higher luminosity and lower temperature than Sun. But they are not big enough like supergiants to have a volume that could include the Sun and inner planets inside.

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.

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​:
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​:
The electron is in ground state, which is $n=1$.
It absorbs 12.09$eV$ of energy, which is the exact difference between $n=1$ and $n=3$ on the diagram. Therefore, the electron will be in $n=3$, which is the second excited state.

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​:
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​:

Moderators slow down the speed of the neutrons (not electrons) so that they can be captured by nuclei of fuel and initiate further fission reactions.

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​:
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.  At the time of this experiment (1911), neither protons nor neutrons had been discovered.

B.  This experiment did not reveal anything about electrons.

C.  This is the correct answer. The conclusion from the alpha particle scattering experiment was that there are high concentrations of positive charge and mass in the middle of an atom, with most of the atom empty space.

D.  This experiment was not specific to gold. Gold foil was used in the experiment but the experiment was about the structure of matter in general. There was also no absorption observed.

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​:

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