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

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

2+A=B+0

Therefore the answer is D.

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​:
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.

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 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​:

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​:
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​:
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​:

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.

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​:
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​:
$Q$ 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​:

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​:

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