Explanation​:

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

Explanation​:
Statement I is true. The nuclear binding energy is the energy released when a nucleus is formed from its constituent nucleons. The final nucleus will have less mass than the separately considered nucleons, and the difference in that mass is known as the mass defect. The mass defect is related to the binding energy of the nucleus by ΔE=Δmc^2.

Statement II is true, as it gives the definition of nuclear binding energy.

Statement III is not true. The energy required to remove a particle from a nucleus is similar to the binding energy per nucleon.

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 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​:
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​:

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

Explanation​:
For one atomic mass unit ($u$), the mass defect is 931.5 MeVc^−2
Total mass defect = $Nu$ = 931.5 NMeVc^−2
This means the binding energy is 931.5 $MeV$

For binding energy per nucleon, dividing by nucleon number A gives 931.5 N/A MeV.
Note that the units of MeVc^−2 are mass units and not energy units.

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 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​:

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.

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​:
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.

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.

Explanation​:
The lengths of the arrows represent the amount of energy released in the transition.

From the formula E=hc/λ​

It is send that the energy of a photon is inversely proportional to its wavelength, therefore the decreasing order of wavelengths will be the increasing order of energies, which is given in C.

The lowest energy transition is shown as λ1​, therefore the λ1​ photon will have the longest wavelength.

λ2​ has the largest energy and therfore will have the shortest wavelength.

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​:

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.