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Transcript of Quantum Physics
Energy E= 10 electronvolt
Basically we're solving for the wavelength
To convert it to Joules multiply by 1.6 x 10^-19
Work: E= 10 eV x (1.6 x 10^-19)
E= hf E= 1.6 x 10^-18
E= hc/w w=(6.63 x 10^-34)*(3.0 x 10^8)/(1.6 x 10^-18)
w=hc/E w= 1.24 x 10^-7 m This is ultraviolet
Planck Constant h= 6.63 x 10^-34 Js
Frequency f= 10^26
E= 6.63 x 10^-34 Js x 10^26 Hz
E= 6.6 x 10^-8 J
2. Visible light has wavelengths in the range 400nm (violet) to 700nm (red). Calculate the energy of a photon of red light and a photon of violet light
Wavelength 1 : 400nm converted to meters = 4.0 x 10^-8 m
Wavelength 2: 700nm converted to meters = 7.0 x 10^-8 m
Speed of light in a vacuum c= 3.0 x 10^8 m/s
Planck constant h= 6.63 x 10^-34 Js
E= (6.63 x 10^-34)(3.0 x 10^8)/7.0 x 10^-8
E=2.84 x 10^-18 J
E= (6.63 x 10^-34)(3.0 x 10^8)/4.0 x 10^-8
E=4.97 x 10^-18 J
5. An electron travels through a cell of e.m.f. 1.2 V. How much energy is transferred to the electron? Give your answer in eV and in J
Voltage v= 1.2 V
Multiply by 1.6 x 10^-19 to get Joules
Divide by 1.6 x 10^-19 to get Electronvolt
E= 1.2eV because it is an electron that with voltage
E= 1.2eV x (1.6 x 10^-19)
E= 1.92 x 10^1-19
7. To which region of the electromagnetic spectrum does a photon of energy 10eV?
The Photoelectric Effect
Work Function Ø = 2.8 x 10^-19
Wavelength w = 2.4 x 10^-7
Mass of electron m = 9.1 x10^-19
a. E= hc/w
E= ((6.63 x 10^-34) x(3.0 x 10^8))/(2.4 x10^-7)
E= 8.29 x 10^-19 J
b. E= Ø + K
K= (8.29 x 10^-19) - (2.8 x 10^-19)
K= 5.49 x 10^-19 J
V= ((2 x (5.49 x 10^-19))/(9.1 x 10^-31))^1/2
V= 1.098 x 10^6 m/s
12. Electromagnetic waves of wavelength 2.4 x 10^-7 m are incident on the surface of a metal whose work function is 2.8 x 10^-19 J.
a. Calculate the energy of a single photon.
b. Calculate the maximum Kinetic energy
c. Determine the maximum speed.
E= 1.0eV , 2.0 eV , 3.0eV
Work function Ø= 1.8eV
Formula: Energy of a photon E= Work function Ø + Kinetic energy K
**For an electron to be released from the metal, it need an about 10^-19 J
a. E= Ø + K
1.0 = 1.8 + K 2.0 = 1.8 + K 3.0 = 1.8 + K
-0.8eV = K 0.2 eV = K 1.2 eV = K
This does not work This works This works
b. K = 0.2 eV x (1.6 x 10^-19) K= 1.2 eV x (1.6 x 10^-19)
K= 3.2 x 10^-20 J K= 1.92 x 10^-19 J
10. Photons of energies 1.0 eV, 2.0 eV and 3.0 eV strike a metal surface whose work function is 1.8 eV.
a. State which of these photons could cause the release of an electron from the metal
b. Calculate the maximum kinetic energies of the electrons released in each case. Give your answers in eV and J.
Particle or Wave?
Nature of Light
When objects give off light, we can disperse this light - essentially, we can split it up. For instance, if you disperse white light, you can see a continuous spectrum of a range of colors all in a row ("continuous")!
The Photoelectric effect
Created by: Micael Joboram, Brandon Galvez, Payton Denny, Tasha Petrik
Test Yourself Questions
1. Calculate the energy of a high-energy y-photon, frequency 10^26 Hz
Electron Waves and Diffraction
Electron diffraction refers to the wave nature of electrons. However, from a technical or practical point of view, it may be regarded as a technique used to study matter by firing electrons at a sample of graphite film and observing the resulting interference pattern. This phenomenon is commonly known as wave–particle duality, which states that a particle of matter (in this case the incident electron) can be described as a wave. For this reason, an electron can be regarded as a wave much like sound or water waves.
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Why should we care care?
This part of physics is called spectroscopy, and it can be used to determine the chemical composition of stars! Ever wondered what Alpha Centauri's made of? You can find out! Really!
...How? That seems really difficult.
Nope! Elements produce different colors, and different colors have different wavelengths. When you view a faraway star (or other hot-gas object!) through a diffraction grating. you will produce what is known as a line spectrum. This * produces patterns using the individual wavelengths of elements, allowing us to look at the spectrum and "read" the chemical composition of the star! Whoa!!
But wait! There's more!
*The line spectrum described above is only one kind: an emission line spectrum! There's also an absorption line spectrum!
The emission line spectrum sounds kind of boring, right? Intermittent lines on some background? That's exactly what it is. But, wow, how cool would it look if we reversed it?! Had the colors be the background and the black be the intermittent lines?! If you think that sounds cool, you're going to love the absorption line spectrum!
...So what exactly is an absorption line spectrum?
Emission line spectra are really useful - when you have a heated gas. But what about a star's cooler gases? A star's interior emits white light, which then passes through the star's cooler outside layers. If the star is emitting white light, we should see the continuous spectrum of all the colors, right? Well, we would, but there's something in the way: those cooler gases. Some of the light emitted by the star's interior is absorbed by the cooler outer gases. This is why we get seemingly "missing" lines on the spectrum. These missing lines correlate directly with the cooler outer gases, telling us the star's composition.
One point of view portrays light as wave-like in nature, producing energy that traverses through space in a manner similar to the ripples spreading across the surface of a still pond after being disturbed by a dropped rock. The opposing view holds that light is composed of a steady stream of particles, much like tiny droplets of water sprayed from a garden hose nozzle. During the past few centuries, the consensus of opinion has wavered with one view prevailing for a period of time, only to be overturned by evidence for the other. Only during the first decades of the twentieth century was enough compelling evidence collected to provide a comprehensive answer, and to everyone's surprise, both theories turned out to be correct, at least in the most part.
Nature of light
When an electron changes its energy from one level to another, it either emits or absorbs a single photon. The energy of the photon hf is equal to the difference in energy between the two levels.
How does it Work?
Glass absorbs the radiation of the light which in turn, prevents energy from the electrons to reach the multimeter.
Different colors allow for different amounts of light because of the smaller and higher frequencies each one possess.
The threshold frequency is defined as the minimum frequency required to release electrons from the source of a metal.
A single electrons minimal energy requirement in order to break from the metal is defined as (Phi).
Einstein's equation for the energy of photons (Photoelectric Equation) is:
How is this even possible??
Electrons can only have certain amounts of energy. We represent these as energy levels. If an element has energy levels of 1 and 2, it can never have a 1.5 energy. When an electron moves from a high energy level to a low energy level, it loses energy - in the form of light. If an element has energy levels of 1 and 2, the photon emitted would have a 1 energy, the difference between 2 and 1. Each element has its own energy levels, and hence its own photon energies, wavelengths, and line spectra.