Explanation:

The decay constant depends on the radioactive material. It does not change for a given nuclide. In this case it is $\lambda$ for both samples as they are the same nuclide. Hence, I is incorrect.

Activity depends on the mass of the sample. If it is $A$ for mass $m$ at $t=0$, it will be $2A$ for $2m$. But after two half lives, it will be halved twice and become A/2. So, II is correct.

After one half life, half of $2m$ decays, and will leave a mass m decayed, while another m is active. After the second half life, half of the remaining $m$ decays leaving just another m/2 active. So the total mass of decayed material is 3m/2. Hence III is incorrect.

Explanation:

A: From his postulates, Bohr showed that the energy of electrons is quantised. Hence, explained why they don't lose energy continuously. This explained why electrons don't spiral into the nucleus. So, A is correct.

B: Bohr's model works only for atoms and ions with single electrons. B is not true. Hence this is the correct choice.

C and D: Based on the assumption that angular momentum of atomic electrons is quantised, Bohr proved that their energies, and orbital sizes are also quantised. So, C and D are correct also.

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:

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 electron is in ground state, which is $n=1$.
It absorbs 12.09$eV$ of energy, which is the exact difference between $n=1$ and $n=3$ on the diagram. Therefore, the electron will be in $n=3$, which is the second excited state.

Explanation:

All photons carry the same energy (corresponding to the frequency of yellow light).

That means the maximum kinetic energy is the same for electrons in both cases. Hence equal stopping voltages.

Hence higher intensity implies larger number of photons per second, and thus larger number of electrons ejected, leading to a higher current. Thus i2>i1

Explanation:

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:
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:

Energy of the incident photon ($E=hf$), goes into:

1. releasing an electron (hfA and hfB respectively), and
2. imparting kinetic energy to the electron. Since stopping voltage converts this kinetic energy into electrical potential energy, this is measured by eVA and eVB for respective materials.

Since the second plate takes more energy to release the electron (∵fB>fA), it gets less kinetic energy after release, and hence smaller stopping potential.

Explanation:
For one atomic mass unit ($u$), the mass defect is 931.5 MeVc^−2
Total mass defect = $Nu$ = 931.5 NMeVc^−2
This means the binding energy is 931.5 $MeV$

For binding energy per nucleon, dividing by nucleon number A gives 931.5 N/A MeV.
Note that the units of MeVc^−2 are mass units and not energy units.

Explanation:
The number of protons on both sides of the equation must be the same.

2+A=B+0

Therefore the answer is D.

Explanation:

In the given reaction, four protons combine to form the nucleus of a helium atom. This process is known as fusion, where smaller nuclei merge to create a larger one. During the annihilation process, two positrons and two electrons interact and transform into gamma rays. Therefore, the energy they stored due to their masses is released. The energy released can be calculated by using the following formula:

• E=mc^2

Since the mass of the positron and electron are the same:

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:
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:
$Q$ 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:

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:
The lengths of the arrows represent the amount of energy released in the transition.

From the formula E=hc/λ

It is send 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.  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:

I is incorrect: Photoelectric effect reflects the particle behavior of light.

II is correct: In the Davisson-Germer experiment electrons were scattered from atoms and these scattered electrons produced an interference pattern just like waves.

III is incorrect: Young's double slit experiment shows wave behavior of light