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

I: CORRECT Decreasing the wavelength is the same as increasing the frequency of the incident radiation. This increases the quantum of photon energy and results in the emission of electrons.

II: INCORRECT Higher intensity of the same light, has no effect on the electrons. Since each photon still carries the same energy as before.

III: CORRECT Reducing the threshold frequency might lead to a point where it is less than the frequency of the incident light (c/λ), and hence result in emission of electrons.

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

2+A=B+0

Therefore the answer is D.

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​:

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​:

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 SI unit of radioactive decay constant is s^−1, which is the same as that of frequency.

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​:
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​:
In the absence of the strong nuclear force there is no force to overcome repulsive electrostatic force between protons. Therefore, more protons in a large nucleus results in more repulsive electrostatic force.

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.

In beta decays, the continuous energy spectra and apparent violation of the law of conservation of momentum let to the postulation of the existence of the neutrino as a particle that could account for the anomaly. Therefore the answer is C.

Nuclear energy levels are evidenced by the discrete energies of alpha particle emissions along with the discrete energies of gamma emissions caused by the nucleus transitioning to lower energy states.

Absorption and emission spectra give evidence of atomic energy levels.

Positrons were first observed in cloud chamber experiments involving cosmic rays.

Explanation​:

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​:
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.

The wave model assumes that the energy from the light is delivered in one continuous wave. As the wavelengths of the lasers are equal, the property that would increase intensity would be the amplitude of the light wave.

In the particle model, light is composed of photons. The photon energy for each laser is equal, so a higher intensity is achieved by when there are more photons emitted per second.

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​:
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 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​:

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