Diffraction occurs when a wave passes into a region behind an obstruction (or spreads out when passing through an aperture).

Going through the first filter, the intensity of initially unpolarized light decreases by 1/2, so . For each filter after the first, the intensity decreases by an additional factor of , where  is the angle between adjacent polarizers. So, after the second filter, we have can see that . Similarly, after the third filter, we have .

The first filter polarizes the light horizontally, only allowing light to pass if it is oscillating horizontally. This would decrease the intensity by one-half.

The second filter would polarize the light vertically, only allowing light to pass if it oscillates vertically. The first filter, however, has already blocked all non-horizontal waves, including any vertical waves; thus, there are no remaining waves that can pass through the vertical filter. The light is fully blocked by this combination.

Polarization is the property that allows tansverse waves to oscillate in multiple orientations. A transverse wave can oscillate, for example, in either the xy-plane or the yz-plane.

Sound waves are longitudinal, and thus do no experience polarization as medium is displaced in one direction only. A longitudinal wave will travel in only one dimension via compression and rarefraction.

We can see that eV differences between the shells for this atom range anywhere from  (between n2 and n1) to  (between n4 and n1). We can use these two values to find the range of wavelengths that can be produced by photons from this atom. To find the wavelength, we will use the formula:

In this equation,  is the energy in Joules,  is Planck's constant,  is the speed of light, and  is the wavelength.

First, convert the energy changes to Joules using the given conversion factor.

Use these energies in the first equation to solve for the range of possible wavelengths.

The visible spectrum spans from approximately to , with smaller wavelengths corresponding to violet and blue and larger wavelengths corresponding to red. Since the range of possible wavelengths only overlaps with a small portion of the visible spectrum, we can see that the only color that can be generated when an electron falls from a higher level to a lower level is violet. The other transitions will generate photons in the ultraviolet range.

Choice 4 is correct because blue light is much more energetic than red light. The most intense low-frequency light will not generate any current at all, because the wavelength is not sufficiently energetic to displace electrons from the photo-electric surface; however, once it is established that a certain frequency is capable of generating current, then the amount of current is dependent upon intensity.

Mnemonic:  “Blue is better.”

Absorbing a photon would have the effect of pushing the atom into a higher energy state, in this case n = 3 or n = 4. Photons with an energy equal to the difference betweeen E2 and E3 or between E2 and E4, could be absorbed.

E3 – E2 = –1.51 – (–3.40) = 1.89eV

E4 – E2 = –0.85 – (–3.40) = 2.55eV

To calculate photon energy from an electron transition, we use the following equation.

In the formula,  is the initial energy level and  is the final energy level. is a constant for the given compound. Our first step will to find the difference described in the formula using the constant given for hydrogen and an estimate for the energy produced.

We can use guess and check to estimate the discrete values that can be used for the electron energy levels.

We find that if the initial energy level is 4 and the final energy level is 1, the value of the difference is approximately -0.94.

An electron transition from energy level 4 to energy level 1 would produce a photon in the appropriate range.

This question can be approached in two ways. The first way is to have a general understanding of the photoelectric effect, and its equation: _, where the kinetic energy of an ejected electron is equal to difference between the energy of the incoming photon, and the work function of the metal plate. If the energy of the photon is greater than that of the work function, an electron will be emitted with a speed that is proportional that difference, thus the greater the energy of the photon, the greater the total kinetic energy, and the faster the speed of the outgoing electron.

Since wavelength is inversely proportional to energy, and because we know that blue light has a wavelength around 400nm, and red light approximately 700nm, we would expect blue light to carry the most energy and thus result in the fastest ejected electron. 

A second approach to this question is to use critical reasoning. Using the concept of energy conservation, we can predict the energy of the incoming photon will be transferred to the outgoing electron. Because we know that energy is inverse to wavelength, the lowest wavelength photon will have the most energy. Applying conservation of energy principles and the fact that energy is directly proportional to velocity, it is a good assumption to reason that the lowest wavelength photon will create the highest velocity electron. This leads to the answer of blue light.

A photon is emitted if the electron goes from a higher to lower energy level, so we need a choice where the energy level, n, decreases. Also, we need to look for the transition that has the smallest energy difference, since frequency is proportional to energy (f  = E/h, where h is Planck's constant). Higher energy levels are closer together, so the highest pair of levels has the smallest difference in energy and the lowest frequency of emitted photons.