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**Chapter 27 Quantum Theory**

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Objectives 27.1 Describe the spectrum emitted by a hot body and explain the basic theory that underlies the emission of hot-body radiation 27.1 Explain the photoelectric effect and recognize that quantum theory can explain it, whereas the wave theory cannot 27.1 Explain the Compton effect and describe it in terms of the momentum and energy of the photon

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Objectives 27.1 Describe the experiments that demonstrate the particle like properties of electromagnetic radiation 27.2 Describe evidence of the wave nature of matter and solve problems relating wavelength to particle momentum 27.2 Recognize the dual nature of both waves and particles and the importance of the Heisenberg uncertainty principle

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**Black Body Radiation The hotter an object is**

The more energy radiated away The higher peak frequency emitted by the object Every object emits electromagnetic waves Result of vibrating particles and electrons falling down

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Quantized Energy The energy an atom has is quantized, meaning it can only have certain energies Atoms in a solid can only vibrate at certain/specific frequencies Energy is emitted when the atom changes its vibration E = nhf Where n is an integer, h is Planck’s constant, and f is frequency. 6.63x 10-34 Energy = 2hf, or 3hf, or 4hf and so on

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Quantized Energy If an atom of energy 4hf were to change its energy to 3hf, it would emit radiation, equal to 1hf. 5hf to 3hf would emit 2hf amount of energy Going up requires receiving certain energies as well These distinct packets of energy we call photons

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Photoelectric Effect The emission of electrons when EM radiation hits an object is known as the photoelectric effect

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Photoelectric Effect Not all radiation results in electrons given off. Need a minimum frequency, called the threshold frequency

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Photoelectric Effect Different materials have different threshold frequencies due to their differences in structure Chem Connection Ionization Energy Electron Orbitals Metals low Non-metals high

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Photoelectric Effect If a red EM wave doesn’t ionize an object, 2red, 3red, or 1000red won’t either. A bulb that produces more light doesn’t make it ionize Need the right frequency

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**Common Uses of Photoelectric Effect**

Garage door openers Doors that open automatically for you when you get close

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Photoelectric Effect Threshold Frequency is the minimum amount of energy needed What happens to the extra energy? Turns into Kinetic Energy of the electron KE = hf – hf0 Kinetic Energy of electron equals the incident frequency minus the threshold frequency Electrons can’t store energy, only get one photon (again, 2reds doesn’t help)

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Photoelectric Effect Not every electron in an atom requires the same amount of energy to escape. Different subshells They will leave with different amounts of KE

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Electron Volt Since electrons are small, and a joule is a very large unit, there is another way to discuss energy of an electron The electron volt (eV) 1 eV = 1.6 x Joules Same as electron charge for convience This is how much energy an electron gets when it is accelerated across a 1.0 volt potential different

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Electron Volt

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**Energy of an EM wave revisited**

E = nhf rearranged for electron volts E = (1240 eV * nm) / wavelength How much energy (in eV) does a photon of wavelength 620 nm have?

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**Stopping Potential How we determined the KE of electrons**

Adjust Voltage until the electrons don’t make it

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Simulator

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Work Function The threshold frequency is related to the energy needed to free the most weakly bound electron from a an object. This is called the work function To find this energy, use E = (1240 eV/nm) / wavelength

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Questions 1) The stopping potential for a photoelectric cell is 5.7 V. Calculate the KE of the emitted photoelectrons in eV. The threshold wavelength of zinc is 310 nm A) Find the threshold frequency of zinc B) What is the work function in eV of zinc? C) Zinc is illuminated by UV of 240 nm. What is the KE of the electron emitted?

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**Answered 1) 5.7 eV is the KE of the emitted electrons**

A) c / wavelength = frequency Frequency = 9.7 x 1014 B) / 310 = 4.0 eV C) (1240 / 240) – (1240 / 310) = 1.2 eV

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Compton Effect Even though photon’s do NOT have mass*, they still have momentum *Energy can be converted to mass technically Momentum of a photon is equal to Planck’s Constant Wavelength

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What it all means The initial photon transfers energy and momentum to the electrons. This transfer of momentum leads to a net loss of energy for the ejected photons (they have a longer wavelength)

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Shifting to waves The last ideas are evidence that photons are particles. We will now discuss wavelike properties of photons

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de Broglie Wavelength Since electrons diffract, de Broglie suggested they were also waves. Since everything is made of electrons/protons, they should all have wavelike characteristics Wavelength (de Broglie) = Planck’s Constant / Momentum of object

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**Wavelength of Baseball**

Wavelength = 6.63 x / (0.25 kg)(20m/s) = 1.3 x 10-34 Too small to have an observable effect What about an electron? Its wave is large enough to be significant compared to itself

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Question An electron has a speed of 5.1 million m/s. What is its de Broglie Wavelength? Momentum = Mass x Velocity P = (9.11 x kg)(5.1 million m/s) h / p = wavelength 6.63 x / 4.6 x 10-24 0.14 nm

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Where are you electron? How do we see where an electron is (it’s location)? By having a photon hit it. But doesn’t the photon move the electron? Yes, so if we know where it is, we have no idea where it is off too

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**Heisenberg Uncertainty Principle**

The more certain you are of location, the less certain you are off the momentum

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Quantum Physics. Black Body Radiation Intensity of blackbody radiation Classical Rayleigh-Jeans law for radiation emission Planck’s expression h = 6.626.

Quantum Physics. Black Body Radiation Intensity of blackbody radiation Classical Rayleigh-Jeans law for radiation emission Planck’s expression h = 6.626.

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