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​:
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.

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.

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​:
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​:
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 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​:

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​:

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​:
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​:
$Q$ 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​:

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.

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​:
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.

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​:
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