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.

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:

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.

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.

The light portion of the electromagnetic spectrum, from lowest to highest frequency, is red, orange, yellow, green, blue, indigo, violet (ROYGBIV).

Frequency is proportional to temperature, and wavelength is inversely proportional to frequency. Since the energy level corresponds with the temperature, objects that emit a higher frequency and shorter wavelength photon will have higher energy. This corresponds with violet, as it is the highest frequency (shortest wavelength) of visible light

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.

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