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:
Energy is released in both nuclear fission and nuclear fusion. Hence the total nuclear mass in both cases decreases, which corresponds to greater mass defect and greater binding energy per nucleon.

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.

Of the choices listed, 32S has a binding energy per nucleon slightly greater than 8 $MeV$. 238U is slightly less than 8 $MeV$. The values for 2H and 4He are much lower.

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.

From the formula E=hc/λ

It is seend 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.

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:

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:
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.

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:
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 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:
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:

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:

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 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:
$Q$ 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:
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.