Explanation​:

The stars have greater mass have higher luminosity which results in a shorter time to run out of hydrogen. Therefore, the lifetime of a star on the main sequence is roughly inversely proportional to the mass of the star. Since the star has greater mass than the sun, time is less than $T$.

Moreover, the mass of the star is less than four times the solar mass. Hence, it is a moderate-mass star and it will become a red giant in its next evolutionary stage.

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​:
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​:
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​:
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​:

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​:
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​:
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.

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​:

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​:
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​:
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.

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​:
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.

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​:
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​:
$Q$ 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​:
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​:

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.