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Lesson Objectives

  • Describe the relationships between speed, wavelength, and frequency of light.
  • Understand the photoelectric effect and how it is related to the wave-particle duality of light.
  • Describe how changes in electron energies lead to atomic emission spectra.
  • Describe the Bohr model of the atom.

Lesson Vocabulary

  • atomic emission spectrum
  • electromagnetic radiation
  • electromagnetic spectrum
  • excited state
  • frequency
  • ground state
  • photoelectric effect
  • photon
  • quantum
  • wavelength

Properties of Light

The nuclear atomic model proposed by Rutherford was a great improvement over previous models, but it was still not complete. It did not fully explain the location and behavior of the electrons in the vast space outside of the nucleus. For example, it was well known that oppositely charged particles attract one another. Rutherford’s model did not explain why the electrons don’t simply move toward and eventually collide with the nucleus. A number of experiments were conducted in the early twentieth century that focused on the ability of matter to absorb and emit light. These studies showed that certain phenomena associated with light revealed a great deal about the nature of matter, energy, and atomic structure.

Wave Nature of Light

In order to begin to understand the nature of the electron, we first need to look at the properties of light. Prior to 1900, scientists thought light behaved solely as a wave. As we will see later, this began to change as new experiments demonstrated that light also has some of the characteristics of a particle. First, we will examine the wavelike properties of light.

Visible light is one type of electromagnetic radiation, which is a form of energy that exhibits wavelike behavior as it moves through space. Other types of electromagnetic radiation include gamma rays, x-rays, ultraviolet light, infrared light, microwaves, and radio waves. Figure below shows the electromagnetic spectrum, which includes all forms of electromagnetic radiation. Notice that visible light makes up only a very, very small portion of the entire electromagnetic spectrum. All electromagnetic radiation moves through a vacuum at a constant speed of 2.998 × 108 m/s. While the presence of air molecules slows the speed of light by a very small amount, we will still use a value of 3.00 × 108 m/s for the speed of light in air.

The electromagnetic spectrum encompasses a very wide range of wavelengths and frequencies. Visible light is only a very small portion of the spectrum, with wavelengths from 400-700 nm.

You can tour the electromagnetic spectrum at http://missionscience.nasa.gov/ems/index.html.

Waves are characterized by their repetitive motion. Imagine a toy boat riding the waves in a wave pool. As the water wave passes under the boat, it moves up and down in a regular and repeated fashion. While the wave travels horizontally, the boat only travels vertically up and down. Figure below shows two examples of waves.

(A) A wave consists of alternation crests and troughs. The wavelength is defined as the distance between any two consecutive identical points on the waveform. The amplitude is the height of the wave. (B) A wave with a short wavelength (top) has a high frequency because more waves pass a given point in a certain amount of time. A wave with a longer wavelength (bottom) has a lower frequency.

A wave cycle consists of one complete wave – starting at the zero point, going up to a wave crest, going back down to a wave trough, and back to the zero point again. The wavelength of a wave is the distance between any two corresponding points on adjacent waves. It is easiest to visualize the wavelength of a wave as the distance from one wave crest to the next. In an equation, wavelength is represented by the Greek letter lambda (\lambda). Depending on the type of wave, wavelength can be measured in meters, centimeters, or nanometers, (1 m = 109 nm). The frequency, represented by the Greek letter nu (\nu), is the number of waves that pass a certain point in a specified amount of time. Typically, frequency is measured in units of cycles per second or waves per second. One wave per second is also called a Hertz (Hz) and in SI units is a reciprocal second (s−1).

Figure above (B) shows an important relationship between the wavelength and frequency of a wave. The top wave clearly has a shorter wavelength than the second wave. However, if you picture yourself at a stationary point watching these waves pass by, more waves of the first kind would pass by in a given amount of time. Thus the frequency of the first waves is greater than that of the second waves. Wavelength and frequency are therefore inversely related. As the wavelength of a wave increases, its frequency decreases. The equation that relates the two is:

c=\lambda \nu

The variable c is the speed of light. For the relationship to hold mathematically, if the speed of light is used in m/s, the wavelength must be in meters and the frequency in Hertz.

Returning to Figure above, you can see how the electromagnetic spectrum displays a wide variation in wavelength and frequency. Radio waves have wavelengths of as long as hundreds of meters, while the wavelength of gamma rays are on the order of 10−12 m. The corresponding frequencies range from 106 to 1021 Hz. Visible light can be split into colors with the use of a prism (Figure below), yielding the visible spectrum of light. Red light has the longest wavelength and lowest frequency, while violet light has the shortest wavelength and highest frequency. Visible light wavelength ranges from about 400 – 700 nm with frequencies in the range of 1014 Hz.

A small beam of white light is (refracted) bent as it passes through a glass prism. The shorter the wavelength of light, the greater is the refraction, so the light is separated into all its colors.

Sample Problem 5.1: Wavelength and Frequency

The color orange within the visible light spectrum has a wavelength of about 620 nm. What is the frequency of orange light?

Step 1: List the known quantities and plan the problem.

Known

  • wavelength (\lambda) = 620 nm
  • speed of light (c) = 3.00 × 108 m/s
  • conversion factor 1 m = 109 nm

Unknown

  • Frequency (\nu)

Convert the wavelength to m, then apply the equation c=\lambda \nu and solve for frequency. Dividing both sides of the equation by \lambda yields:

\nu = \dfrac{c}{\lambda}

Step 2: Calculate

& 620 \ \text{nm} \times \left ( \dfrac{1 \ \text{m}} {10^9 \ \text{nm}} \right ) = 6.20 \times 10^{-7} \ \text{m} \\& \nu = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8 \ \text{m/s}} {6.20 \times 10^{-7} \ \text{m}} = 4.8 \times 10^{14} \ \text{Hz}

Step 3: Think about your result.

The value for the frequency falls within the range for visible light.

Practice Problems
  1. What is the frequency of radiation of wavelength 2.7 × 10−9 m? In what region of the electromagnetic spectrum is this radiation?
  2. Calculate the wavelength in nm of visible light with a frequency of 6.80 × 1014 Hz.

Quantum Physics

German physicist Max Planck (1858-1947) studied the emission of light by hot objects. You have likely seen a heated metal object glow an orange-red color (Figure below).

A heated object may glow different colors. The atoms in this piece of metal are releasing energy in discrete units called quanta.

Classical physics, which explains the behavior of large, everyday objects, predicted that a hot object would emit electromagnetic energy in a continuous fashion. In other words, every wavelength of light could possibly be emitted. Instead, what Planck found by analyzing the spectra was that the energy of the hot body could only be lost in small discrete units. A quantum is the minimum quantity of energy that can either be lost or gained by an atom. An analogy is that a brick wall can only undergo a change in height by units of one or more bricks and not by any possible height. Planck showed that the amount of radiant energy absorbed or emitted by an object is directly proportional to the frequency of the radiation.

E = h \nu

In the equation, E is the energy, in joules, of a quantum of radiation, \nu is the frequency, and h is a fundamental constant called Planck’s constant. The value of Planck’s constant is h = 6.626 × 10−34 J•s. The energy of any system must increase or decrease in units of h \nu. A small energy change results in the emission or absorption of low-frequency radiation, while a large energy change results in the emission or absorption of high-frequency radiation.

The Photoelectric Effect and the Particle Nature of Light

In 1905 Albert Einstein (1879-1955) proposed that light be described as quanta of energy that behave as particles. A photon is a particle of electromagnetic radiation that has zero mass and carries a quantum of energy. The energy of photons of light is quantized according to the E=h \nu equation. For many years light had been described using only wave concepts, and scientists trained in classical physics found this wave-particle duality of light to be a difficult idea to accept. A key experiment that was explained by Einstein using light’s particle nature was called the photoelectric effect.

The photoelectric effect is a phenomenon that occurs when light shined onto a metal surface causes the ejection of electrons from that metal. It was observed that only certain frequencies of light are able to cause the ejection of electrons. If the frequency of the incident light is too low (red light, for example), then no electrons were ejected even if the intensity of the light was very high or it was shone onto the surface for a long time. If the frequency of the light was higher (green light, for example), then electrons were able to be ejected from the metal surface even if the intensity of the light was very low or it was shone for only a short time. This minimum frequency needed to cause electron ejection is referred to as the threshold frequency.

Classical physics was unable to explain the photoelectric effect. If classical physics applied to this situation, the electron in the metal could eventually collect enough energy to be ejected from the surface even if the incoming light was of low frequency. Einstein used the particle theory of light to explain the photoelectric effect as shown in Figure below.

Low frequency light (red) is unable to cause ejection of electrons from the metal surface. At or above the threshold frequency (green) electrons are ejected. Even higher frequency incoming light (blue) causes ejection of the same number of electrons but with greater speed.

Consider the E = h \nu equation. The E is the minimum energy that is required in order for the metal’s electron to be ejected. If the incoming light’s frequency, \nu, is below the threshold frequency, there will never be enough energy to cause electron to be ejected. If the frequency is equal to or higher than the threshold frequency, electrons will be ejected. As the frequency increases beyond the threshold, the ejected electrons simply move faster. An increase in the intensity of incoming light that is above the threshold frequency causes the number of electrons that are ejected to increase, but they do not travel any faster. The photoelectric effect is applied in devices called photoelectric cells, which are commonly found in everyday items such as a calculator which uses the energy of light to generate electricity (Figure below).

Photoelectric cells convert light energy into electrical energy which powers this calculator.

Run a simulation of the photoelectric effect at http://phet.colorado.edu/en/simulation/photoelectric.

Sample Problem 5.2: Quantized Energy

What is the energy of a photon of green light with a frequency of 5.75 × 1014 Hz?

Step 1: List the known quantities and plan the problem.

Known

  • frequency (\nu) = 5.75 × 1014 Hz
  • Planck’s constant (h) = 6.626 × 10-34 J•s

Unknown

  • energy (E)

Apply the equation E = h \nu to solve for the energy.

Step 2: Calculate

E = (6.626 \times 10^{-34} \ \text{J} \cdot \text{s}) \times (5.75 \times 10^{14} \ \text{Hz}) = 3.81 \times 10^{-19} \ \text{J}

Step 3: Think about your result.

While the resulting energy may seem very small, this is for only one photon of light. Visible quantities of light consist of huge quantities of photons. Recall that a hertz is equal to a reciprocal second, so the units agree in the equation.

Practice Problem
  1. A certain photon of radiation has an energy of 8.72 × 10−21 J. Calculate the frequency and wavelength (in m) of this radiation.

Atomic Emission Spectra

The electrons in an atom tend to be arranged in such a way that the energy of the atom is as low as possible. The ground state of an atom is the lowest energy state of the atom. When those atoms are given energy, the electrons absorb the energy and move to a higher energy level. These energy levels of the electrons in atoms are quantized, meaning again that the electron must move from one energy level to another in discrete steps rather than continuously. An excited state of an atom is a state where its potential energy is higher than the ground state. An atom in the excited state is not stable. When it returns back to the ground state, it releases the energy that it had previously gained in the form of electromagnetic radiation.

So how do atoms gain energy in the first place? One way is to pass an electric current through an enclosed sample of a gas at low pressure. Since the electron energy levels are unique for each element, every gas discharge tube will glow with a distinctive color depending on the identity of the gas (Figure below).

Gas discharge tubes are enclosed glass tubes filled with a gas at low pressure through which an electric current is passed. Electrons in the gaseous atoms first become excited, and then fall back to lower energy levels, emitting light of a distinctive color in the process. Shown are gas discharge tubes of helium, neon, argon, krypton, and xenon.

“Neon” signs are familiar examples of gas discharge tubes. However, only signs that glow with the red-orange color seen in the figure are actually filled with neon. Signs of other colors contain different gases or mixtures of gases.

Scientists studied the distinctive pink color of the gas discharge created by hydrogen gas. When a narrow beam of this light was viewed through a prism, the light was separated into four lines of very specific wavelengths (and frequencies since \lambda and \nu are inversely related). An atomic emission spectrum is the pattern of lines formed when light passes through a prism to separate it into the different frequencies of light it contains. Figure below shows the atomic emission spectrum of hydrogen.

When light from a hydrogen gas discharge tube is passed through a prism, the light is split into four visible lines. Each of these spectral lines corresponds to a different electron transition from a higher energy state to a lower energy state. Every element has a unique atomic emission spectrum, as shown by the examples of helium (He) and iron (Fe).

The four visible lines of hydrogen’s atomic emission spectrum with corresponding wavelengths are violet (410 nm), blue (434 nm), blue-green (486 nm), and red (656 nm). The pinkish color that our eyes see is a combination of these four colors. Every element has its own unique atomic emission spectrum which can be used to identify the gas, a technique which is used extensively in astronomy to identify the composition of distant stars.

Classical theory was unable to explain the existence of atomic emission spectra, also known as line-emission spectra. According to classical physics, a ground state atom would be able to absorb any amount of energy rather than only discrete amounts. Likewise, when the atoms relaxed back to a lower energy state, any amount of energy could be released. This would result in what is known a continuous spectrum, where all wavelengths and frequencies are represented. White light viewed through a prism and a rainbow are examples of continuous spectra. Atomic emission spectra were more proof of the quantized nature of light and led to a new model of the atom based on quantum theory.

Work with simulated discharge lamps at http://phet.colorado.edu/en/simulation/discharge-lamps.

The following video shows a lab demonstration of the atomic emission spectra: http://www.youtube.com/watch?v=955snB6HLB4 (1:39).

Bohr Model of the Atom

Figure below helps explain the process that occurs when an atom is excited and relaxes back to a lower energy.

As an excited atom with energy equal to E2 falls back down to energy E1, it releases energy in the form of a photon of electromagnetic energy. The energy of the photon is Ephoton = E2E1 = .

The energy states of an atom are indicated by E_1 and E_2, with E_2 being higher in energy. As the atom moves from the E_2 excited state down to the lower energy E_1 state, it loses energy by emitting a photon of radiation. The energy of that photon (E_{photon} = h \nu) is equal to the gap in energy between the two states or E_2 - E_1. The observation that the atomic emission spectrum of hydrogen consists of only specific frequencies of light indicates that the possible energy states of the hydrogen atom are fixed. This suggested that the electrons in a hydrogen atom were arranged into specific energy levels.

In 1913 Danish physicist Neils Bohr (1885-1962) proposed a model of the atom that explained the hydrogen atomic emission spectrum. According to the Bohr model, often referred to as a planetary model, the electrons encircle the nucleus of the atom in specific allowable paths called orbits. When the electron is in one of these orbits, its energy is fixed. The ground state of the hydrogen atom, where its energy is lowest, is when the electron is in the orbit that is closest to the nucleus. The orbits that are further from the nucleus are all of successively greater energy. The electron is not allowed to occupy any of the spaces in between the orbits. An everyday analogy to the Bohr model is the rungs of a ladder. As you move up or down a ladder, you can only occupy specific rungs and cannot be in the spaces in between rungs. Moving up the ladder increases your potential energy, while moving down the ladder decreases your energy.

Bohr’s model explains the spectral lines of the hydrogen atomic emission spectrum. While the electron of the atom remains in the ground state, its energy is unchanged. When the atom absorbs one or more quanta of energy, the electron moves from the ground state orbit to an excited state orbit that is further away. Energy levels are designated with the variable n. The ground state is n = 1, the first excited state is n = 2, and so on. The energy that is gained by the atom is equal to the difference in energy between the two energy levels. When the atom relaxes back to a lower energy state, it releases energy that is again equal to the difference in energy of the two orbits (Figure below).

Bohr model of the atom: electron is shown transitioning from the n = 3 energy level to the n = 2 energy level. The photon of light that is emitted has a frequency that corresponds to the difference in energy between the two levels.

The change in energy, \Delta E, then translates to light of a particular frequency being emitted according to the equation \Delta E = h \nu. Recall that the atomic emission spectrum of hydrogen had spectral lines consisting of four different frequencies. This is explained in the Bohr model by the realization that the electron orbits are not equally spaced. As the energy increases further and further from the nucleus, the spacing between the levels gets smaller and smaller.

Based on the wavelengths of the spectral lines, Bohr was able to calculate the energies that the hydrogen electron would have in each of its allowed energy levels. He then mathematically showed which energy level transitions corresponded to the spectral lines in the atomic emission spectrum (Figure below).

The electron energy level diagram for the hydrogen atom shows the electron transitions for the Lyman, Balmer, Paschen, and Brackett series. Bohr’s atomic model mathematically accounted for the atomic emission spectrum of hydrogen.

He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). This is called the Balmer series. Transitions ending in the ground state (n = 1) are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. The transitions called the Paschen series and the Brackett series both result in spectral lines in the infrared region because the energies are too small.

Watch a simulation of the Bohr model of the hydrogen atom at http://www.dlt.ncssm.edu/core/Chapter8-Atomic_Str_Part2/chapter8-Animations/ElectronOrbits.html.

Sample Problem 5.3: Spectral Lines of the Hydrogen Atom

In the hydrogen atom, the change in energy (\Delta E) for the n = 4 to n = 2 electron transition is equal to 4.09 × 10-19 J. Calculate the wavelength (in nm) of the spectral line that results from this electron transition and identify its color.

Step 1: List the known quantities and plan the problem.

Known

  • \Delta E = 4.09 × 10-19 J
  • Planck's constant (h) = 6.626 × 10-34 J•s
  • speed of light (c) = 3.00 × 108 m/s
  • conversion factor 1 m = 109 nm

Unknown

  • frequency (\nu)
  • wavelength (\lambda)

Apply the equation \Delta E = h \nu to solve for the frequency of the emitted light. Then, use the equation c = \lambda \nu to solve for the wavelength in m. Convert to nm.

Step 2: Calculate

& \nu = \frac{\Delta E}{h} = \frac{4.09 \times 10^{-19} \ \text{J}}{6.626 \times 10^{-24} \ \text{J} \cdot \ \text{s}} = 6.17 \times 10^{14} \ \text{Hz} \\& \lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8 \ \text{m/s}} {6.17 \times 10^{14} \ \text{Hz}} = 4.86 \times 10^{-7} \ \text{m} \\& 4.86 \times 10^{-7} \ \text{m} \times \left ( \frac{10^9 \ \text{nm}}{1 \ \text{m}} \right ) = 486 \ \text{nm}

Step 3: Think about your result.

The 486 nm spectral line corresponds to a blue-green color. Then n = 3 to n = 2 transition results in a smaller energy release. This in turn yields a lower frequency and the longer wavelength red spectral line. The n = 5 to n = 2 and n = 6 to n = 2 transitions yield the shorter wavelength lines.

Practice Problems
  1. The energy change (\Delta E) for the n = 2 to n = 1 transition of the Lyman series is 1.64 × 10-18 J. Calculate the wavelength of the resulting spectral line.
  2. The visible red spectral line that results from the n = 3 to n = 2 transition of the hydrogen atom is of wavelength 656 nm. Calculate the energy change that produces this spectral line.

Bohr’s model was a tremendous success in explaining the spectrum of the hydrogen atom. Unfortunately, when the mathematics of the model was applied to atoms with more than one electron, it was not able to correctly predict the frequencies of the spectral lines. While Bohr’s model represented a great advancement in the atomic model and the concept of electron transitions between energy levels is valid, improvements were needed in order to fully understand all atoms and their chemical behavior.

Lesson Summary

  • Light is one part of the entire electromagnetic spectrum and its wave nature can be described by its wavelength and frequency.
  • The particle nature of light is illustrated by the photoelectric effect and atomic emission spectra.
  • Quantum theory states that the electrons of an atom can only exist at discrete energy levels. When electrons transition from a high energy level to a low energy level, energy is released as electromagnetic radiation.
  • In the Bohr model of the atom, electrons are allowed to be only at specific distances from the nucleus called orbits. Each orbit has a characteristic energy level.

Lesson Review Questions

Reviewing Concept

  1. Answer the following:
    1. List five examples of electromagnetic radiation.
    2. What is the speed of all forms of electromagnetic radiations?
    3. List the colors of the visible spectrum in order from shortest to longest wavelength.
  2. Answer the following:
    1. How are the wavelength and frequency of light related?
    2. How are the energy and frequency of light related?
    3. How are the energy and wavelength of light related?
  3. Consider the following regions of the electromagnetic spectrum: (i) x-ray, (ii) infrared, (iii) microwave, (iv) visible, (v) radio wave, (vi) gamma ray, (vii) ultraviolet.
    1. Arrange them in order of increasing wavelength.
    2. Arrange them in order of increasing energy.
  4. Answer the following:
    1. What is a quantum of energy?
    2. Explain how the stacking of poker chips is related to quantum theory.
  5. What happens when a hydrogen atom absorbs a quantum of energy? Use the terms ground state and excited state in your answer.
  6. Explain the difference between a continuous spectrum and an atomic emission spectrum.
  7. Answer the following:
    1. What is the photoelectric effect?
    2. Why does the photoelectric effect support the idea that light can behave as a particle?
  8. Where are electrons located according to the Bohr model?
  9. Use the Bohr model to explain how the atomic emission spectrum of hydrogen is produced.
  10. Answer the following:
    1. Which series in hydrogen’s atomic emission spectrum is composed of visible light?
    2. What are the similarities of each line of that series?

Problems

  1. What is the frequency of electromagnetic radiation that has a wavelength of 0.325 cm? In what region of the electromagnetic spectrum is this radiation located?
  2. What is the frequency of a photon of light that has an energy of 2.95 × 10−17 J?
  3. Answer the following:
    1. Using the equations E = h \nu and c = \lambda \nu, derive an equation that expresses E as a function of \lambda.
    2. Use your equation from part a to determine the energy of a photon of yellow light that has a wavelength of 579 nm.
  4. The change in energy of a certain electron transition in the Paschen series is 1.82 × 10−19 J. Calculate the wavelength (in nm) of the spectral line produced by this transition.

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