Light - MCAT Chemical and Physical Foundations of Biological Systems

The energy of a wave increases with increasing frequency and decreasing wavelength. Considering these different waves, radiowaves possess the longest wavelengths and gamma rays the shortest wavelength, thus gamma rays carry the greatest amount of energy.

Incandescent light bulbs produce visible light of all wavelengths. The mix of red, orange, yellow, green, blue, indigo, and violet (ROYGBIV) give the light its characteristic white appearance. For the MCAT, it is important to know the relative wavelengths of light for the visible spectrum (390 – 700nm) and where visiable wavelengths fit into the overall spectrum of electromagnetic radiation. From longest wavelength to shortest, the sequence of wavelengths is listed below.

Radio > Microwaves > Infrared > Visible > Ultraviolet > X-Rays > Gamma Rays

We know that the visible spectrum has wavelengths of 390nm to 700nm. The wavelengths of light around the 700nm range are red.

Using the ROYGBIV acronym, we know that red has the longest wavelength and violet the shortest. While this may seen to be a difficult question asking for the wavelengths of light that correspond to certain colors, this is helpful on the MCAT for estimating answers and may be worth the time learning. The expected range of wavelengths for the red portion of the visible spectrum is 650nm to 700 nm. Keep in mind that, because it has the longest wavelength, red light will also have the lowest frequency.

The highest energy of any visible light belongs to violet. The greater the wavelength, the lower the energy of the light. The greater the frequency, the higher the energy of the light. This is why ultraviolet light ("ultra" meaning "beyond" violet) is so damaging to DNA. Out of the answer choices, blue light has the lowest wavelength and greatest frequency, making it the highest energy.

Choice 5 is correct. Highly energetic frequencies such as X-rays and gamma rays can displace electrons from materials upon which they impinge. Microwaves are a form of radio waves, which are long-wavelength, low frequency waves with little energy.

Mnemonic: Microwave ovens are fundamentally safe household items.

As Einstein determined, light has properties of both a particle and a wave. In the particle sense, it has mass, velocity, and momentum. The wave property of light allows for diffraction, and constructive and destructive interference. For the MCAT, it is important to know that the photon (the particle of light) has both particle and wave properties. In fact, all objects have both particle and wave properties; however, their wave property becomes less obvious with increasing mass.

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.”

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.

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

This question asks us about the particle nature of light and how much energy a photon would contain. From our light equations, we know that , where E is the energy of a single photon, h is Plank’s constant, and f is the frequency of the photon.

From the information in the problem, we need to determine the frequency.

We need to determine the velocity of the light in the solution. We can use the definition of index of refraction to determine this value, along with the speed of light and the index of refraction of the solution.

Now we can compute the frequency.

Substituting the frequency we found, along with Plank’s constant, we can find the energy.

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.

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.

Work function is the minimum amount of energy, in electron-volts, required to remove an electron from a metal surface. To calculate the energy of a single photon of light, one must use the equation:

In this formula, is energy in Joules, is Planck's Constant, and is the speed of light.

To convert from Joules to eV, one must use the given conversion factor.

Use this value in the energy formula to find the wavelength of the light.

The visible spectrum spans from approximately to , with smaller wavelengths corresponding to violet and blue and larger wavelengths corresponding to red. Even without knowing the exact wavelength correlations in the spectrum, we know that our wavelength is very small and will be found in the violet end of the spectrum.

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.

The energy of a single photon is given by the equation:

We are given the frequency and the value of the constant, allowing us to solve.

The above gives the energy contained in one photon. Next, solve for the energy contained in using Avogadro's number:

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.

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.

The energy of a wave increases with increasing frequency and decreasing wavelength. Considering these different waves, radiowaves possess the longest wavelengths and gamma rays the shortest wavelength, thus gamma rays carry the greatest amount of energy.