Supplemental Activities. Module: Atomic Theory. Section: Electromagnetic Radiation and Matter - Key

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1 Supplemental Activities Module: Atomic Theory Section: Electromagnetic Radiation and Matter - Key

2 Introduction to Electromagnetic Radiation Activity 1 1. What are the two components that make up electromagnetic radiation? Electromagnetic radiation is the result of an oscillating magnetic field perpendicular to an oscillating electric field. 2. Please match the symbol on the left to the appropriate description on the left. c. E a. wavelength, m a. λ e. ν d. h b. the speed of light in a vacuum, 3.00 x 8 m/s c. energy, J d. Planck s constant, x 34 Js b. c e. frequency, Hz or s 1 3. Please write at least two valid equations that use the symbols listen in question 2 above. E = hν and c = λ ν Combining these two equations gives an alternate way to calculate energy with wavelength instead of frequency: E = hc λ Activity 2 1. List the regions of electromagnetic radiation in decreasing order from highest energy to lowest energy. gamma rays > X-Rays > ultraviolet > visible > infrared > microwaves > radio

3 2. What is the range of wavelengths for visible light? Range of frequencies? Wavelengths: (750 nm 400 nm) Frequencies: (4x x 14 Hz) 3. A hypothetical wave has a frequency of 1.0 x 15 Hz. What is its hypothetical, approximate wavelength? State in which region of light this wave exists. c = λν m s = λ( s ) λ = m s s λ = m This would be found in the UV region of the EM spectrum. 4. A hypothetical wave has a wavelength of 5 m. What is its hypothetical, approximate frequency? State in which region of light this wave exists. c = λv 3.00 m s m s λ = m 13 λ = 3.0 Hz 8 = (1.0 5 m) ν This would be found in the Infrared region of the EM spectrum. 5. A hypothetical wave has an energy of JJ. What is its hypothetical, approximate frequency? State in which region of light this wave exists. E = hc v J v = = Hz 34 Js *3.00 This would be found in the Microwave region of the EM spectrum. 6. Electrons are negatively charged and so are attracted to the positive side of a static electric field. In an oscillating electric field, electrons also oscillate (or vibrate). 7. How would doubling the frequency (v, pronounced nu, not vee) of a given wave affect the following values? v 8 m s

4 a. λ It would be halved b. c No change c. E It would be doubled d. h No change 8. Without calculations, rank the frequency of the following waves from highest to lowest frequency: waves with wavelengths of 300 nm, 1 m, 1 km, and 300 Å. Frequency and wavelength are inversely proportional. The shortest wavelengths will thus have the highest frequencies. The ranking should be: 300 Å, 300 nm, 1 m, 1 km. Note: 1 Å = 1 mm Blackbody Radiation and Ultraviolet Catastrophe Activity 1 1. Classical mechanics made a prediction about the frequencies of light emitted by black body radiators at higher and higher temperatures. What was this prediction and did it match the experimental observations? According to classical theory, higher and higher temperatures, radiators would emit light of ever-increasing energy. The math behind the classical theories predicted that radiators would produce extremely high-energy radiation at significant temperatures. However, experimental data proved these predictions false. 2. The scientist Max Planck suggested a solution to this problem dubbed the Ultraviolet Catastrophe. 3. This scientist s suggestion indicated that the emission of a particular frequency requires an oscillator to have a minimum/specific energy. Since physical objects have a limited number of oscillation energies, they have a limited number of emitted frequencies. The Photoelectric Effect

5 Activity 1 Activity 3 1. Draw a picture illustrating the photoelectric effect. Describe the figure and explain how frequency and work function (Φ) relate to the kinetic energy of the emitted electron. 1 photon from the light source corresponds to 1electron ejected if and only if the energy of the photon, h is greater than the work function of the metal, Φ. Nothing happens if h < Φ. Even if you make the light source really bright by increasing the intensity and therefore have more photons hitting the metal, no electrons will be ejected as long as h < Φ. If h > Φ, the kinetic energy of the ejected electron increases as the energy difference between h and Φ increases. 2. You shine 500 nm light on a metal and electrons come off. What will happen if you shine 400 nm light of the same intensity on the metal? a. fewer electrons will come off but with the same velocities b. nothing c. electrons will come off with higher velocities d. the two situations will be identical 3. Explain your answer choice in question 2 above. Devise experiments that would produce the other three outcomes that you did not choose. Using higher energy light (lower wavelength) means that more energy is transferred to the electrons as kinetic energy. In order to produce answer choice a as a result, we would need to maintain a 500 nm light source but lower the intensity. The energy of the light dictates the kinetic energy of the electrons. Intensity affects the number of electrons emitted. In order to produce answer choice b as a result, we would need to increase the wavelength of light until the frequency of light is below the necessary work function of the metal being used.

6 In order to produce answer choice d as a result, we would have to maintain the same setup and not change the wavelength, intensity or type of metal. Activity 4 1. Describe the relationship between frequency of incident light and the kinetic energy of ejected electrons. The kinetic energy of a sodium electron increases linearly with increasing frequency of incident light. However, the frequency of incident light must be above a certain threshold frequency Φ in order for the electron to be ejected. This is the work function of a metal. Each metal has a different work function based on its own particular properties. 2. A packet of quantized energy associated with electromagnetic radiation is called a photon. 3. Famous scientist Einstein proposed the idea of a photon after interpreting the results of the photoelectric effect. He noted that increasing the intensity of light corresponded to more electrons being ejected. However, if no electrons were being ejected in the first place, nothing would happen even if the intensity were increased. This led him to develop the idea of particles of light rather than waves of light energy that interact individually with electrons. Activity 5 1. A photon with enough energy, 5.1 electron volts (ev) of energy to be precise, will eject an electron from a piece of gold. What frequency and wavelength does light with this energy have? Note: 1 ev = 1.60 x 19 joules. E = hν E = 5.1eV J 1eV = J ν = E h ν = J Js = Hz c = λν λ = c ν λ = m s s = m

7 2. Recalling that photon of 5.1 electron volts (ev) of energy will eject an electron from a piece of gold, what would happen if you were to shine a light of 6.5 ev on the gold surface? How is this the same or different from using light of 3.0 ev? What if the metal was Cesium (Φ = 2.1 ev) or Platinum (Φ = 6.35 ev) instead? 6.5 ev is greater than the work function of gold (5.1 ev), so the electron would be ejected from the gold surface. The electrons that are ejected would be faster than electrons ejected while using 5.1 ev light. Light of 3.0 ev cannot overcome the work function for gold, so no electron would be ejected. For Cs, both 6.5 ev and 3.0 ev would eject an electron because the work function is so low. Again, electrons would be faster when the 6.5 ev light is used compared to the 3.0 ev light. For platinum only the 6.5 ev light will be enough to eject an electron. When the 6.5 ev light is being used, the electrons from Cs will be fastest, the electrons from Au will have intermediate speeds and the electrons from Pt will be the slowest. Wave-Particle Duality and Uncertainty Activity 1 1. Young s double slit experiment and the observation of diffraction patterns suggest that light behaves as a wave. 2. The ultraviolet catastrophe outcomes and the photoelectric effect suggest that light behaves as a particle. 3. What does the phrase wave-particle duality mean in your own words? Light has both wavelike and particlelike behaviors. We use the term wave-particle duality to describe this dual nature of light. Activity 2 1. What did Louis de Broglie propose about matter both large and small? de Broglie proposed that all matter exhibits wavelike and particlelike behaviors. Larger, slower objects have lower wavelengths than smaller, faster objects. 2. What equation describes his idea? Please indicate what each variable means (include units).

8 λ = h mv The variable λ is wavelength in meters (m). The constant h is Planck s constant in jouleseconds (J s). The variable m is mass in kilograms (kg). The variable v is velocity in meters per second (m/s). 3. Please calculate the wavelength of a neutron traveling at the speed of light (in a vacuum). c = m s v = c = m s h = J s m neutron = kg Activity 3 λ = h mv J s λ = ( kg)(3.0 8 m s ) λ = m 1. Briefly explain Heisenberg s uncertainty principle in your own words. The uncertainty principle tells us that the location of an object and its trajectory/path/velocity/momentum cannot both be known at the same moment in time. 2. What equation describes this principle? Please indicate what each variable means (include units). x p h 4π The variable x is the uncertainty in the position of an object in meters (m). The variable p is uncertainty in the momentum of an object in kilogram-meters per second (kg m/s). The constant h is Planck s constant. The constant π is pi. 3. Assume that an atom is approximately 0.4nm in diameter and that this value is the uncertainty in position for an electron in an atom. What would the uncertainty in momentum be for this electron? What would the uncertainty in velocity be for this electron?

9 x p h 4π x = 0.4nm = 4 m m electron = kg h = J s x p h 4π h p 4π x p J s 4π (4 m) p kg m s p = m v m v kg m s v kg m s m v kg m s kg v m s

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