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​:

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.  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​:
The energy difference between the ground state and the first excited state of anelectron is given by

−5.5−(−10.4)=4.9 eV

In order to have emitted photons with an energy of 4.9$eV$, electrons must be excited. Incident electrons can excite electrons in orbits with more than sufficient energy while photons need to have the exact amount of energy for moving an electron from one state to another. Therefore a photon with an energy of 5.1$eV$ cannot excite the electrons but others can. So, the answer is C.

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 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​:
For each half-life, half of the remaining nuclides in the sample decay to a different nuclide. In this situation, three half lives occur in the 90 minutes, so only one eighth of the original isotope remains. Because the daughter nuclides are also solid, however, they remain in the sample.

The decay products of beta decay include the daughter nuclide, the beta particle (an electron), and an anti-neutrino. The mass of the electron and the anti-neutrino are negligible compared to the parent nuclide. Because the daughter nuclides contain the same number of nucleons as the parent nuclides, there is effectively no change in mass of the sample.

Explanation​:
The alpha, beta and gamma radiation have different penetrating powers, where alpha is the one with the least penetrating power and won’t pass through the paper sheet, while gamma and beta radiations will pass through the paper sheet.

When entering the magnetic field, gamma radiation will continue in its path without deflection because they have no charge. On the other hand, beta particles have negative charges, so they will be affected by a magnetic force that causes deflection in their path.

To determine the direction of the magnetic force, the right hand rule is used keeping in mind that the velocity direction in the right hand rule is that of a positive charge. So we have the direction of velocity is to the left, the magnetic field is into page, thus the magnetic force will be directed downwards.

*Any other hand-rule convention can also be used.

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​:
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 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 a fission reaction a heavy nucleus splits into two lighter nuclei to release energy. So the total binding energy of daughter nuclei ($Y$ and $Z$) is greater than the parent nucleus ($X$).

From the graph, it can be seen that the binding energy per nucleon of $Y$ (or $Z$) is greater than binding energy per nucleon of $X$

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​:

Since the mass of the star is greater than the Sun, its luminosity should be higher. Therefore, its position on the main sequence should be A instead of C, where C represents the Sun. Moreover, the last stage of the evolution should be white dwarf which is labelled as G.

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 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​:

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.