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:
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 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:

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:

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:

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

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 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:

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

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 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:

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:
Binding energy of $X$; (A)(E)=AE
By using E=mc^2, mass defect; m=AE/c^2
Therefore the answer is A.

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:

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:

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.