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​:
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.  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​:
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​:
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​:
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​:

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​:

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.

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.

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 $X$

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

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​:

To initiate hydrogen fusion, the protons in two hydrogen nuclei must be brought within the range of the strong nuclear force. This can be achieved by raising the temperature of the nuclei so that their kinetic energy is high enough. However, even if the nuclei are close enough, the majority of collisions will not result in fusion.

Therefore, the number of collisions must be high enough to ensure that a significant number of fusion reactions occur. This can be achieved by increasing the density and pressure of the nuclei.

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​:
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.