LE 2.05: Photoelectric Effect Simulation + LE 2.06: Einstein's Nobel Prize
Can the wave model of light explain what you saw happen in the photoelectric effect simulation?
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Did all of the metals have the same threshold wavelength for ejecting electrons from the metal surface?
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Energy, wavelength, frequency triangle
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If a photon of wavelength 375 nm strikes a magnesium surface (φ = 3.68 eV), will an electron be ejected? If yes, what will the kinetic energy of the electron be (in joules)?
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If a photon with a wavelength of 420 nm strikes a potassium surface in a vacuum, will an electron be ejected? If so, what will the electron's kinetic energy be? φ = 3.67 × 10-19 J h = 6.626 × 10-34 J·s c = 3.0 × 108 m/s
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In Step 6 of the procedure, what was the threshold wavelength for potassium (the longest wavelength that still ejected electrons), in nm? (see photo)
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The work function φ for chromium metal is 7.0 x 10-19 J. If an electron on the surface of chromium absorbs a 210 nm ultraviolet photon, how much kinetic energy will the escaped electron have? A) 9.45 x 10-19 J B) 7.0 x 10-19 J C) 2.45 x 10-19 J
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The work function φ is often reported in units of electron-volts, or eV for short. Electron-volts is an energy unit like joules, but one electron-volt is much smaller than one joule: 1 eV = 1.60 x 10-19 J. Therefore the work functions of common metals are typically in the range of 2 - 5 eV, which are much easier numbers to work with than values on the order of 10-19 J. Planck's constant is in units of Js, so if you are given energy values in eV or asked to report an answer in eV, be careful not to mix energy units during the calculation. What frequency of light is necessary to eject an electron with a kinetic energy of 1.30 eV from an aluminum surface (φ = 4.08 eV)?
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When you changed the wavelength from 500 nm to 440 nm in Step 5 of the experiment you made the wavelength of the light shorter and the frequency higher. What effect did this have on the ejected electrons? Select each effect you observed in the simulation when you changed the wavelength of the light. (see photo for answer and explanation of answers)
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Why waves/The wave theory of light fails to explain why Increasing the brightness of the light ejects more electrons, but the electrons have the same kinetic energy as before
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Photoelectric effect equations
(see photo) 2 formats: hv = ϕ + KEe KEe = hv - ϕ KEe = leftover energy from photon hv = energy of photon where v = frequency and h = Planck's constant aka (6.62.6 x 10^-34 J x s) ϕ = energy required to eject an electron (aka e^-) from the metal Remember if dealing with velocity: KE = 1/2mv^2 where m = mass of an electron
photoelectric effect
1) Discovered by Heinrich Hertz 2) Near the end of 1800s an unexplained phenomenon arose called this 3) Shining light (usually violet or ultraviolet light) onto a metal surface in a vacuum could cause electrons to be ejected from the metal surface 4) But the effects they saw couldn't be explained. Light waves would not do what this light was doing.
4 different types of problems involving photoelectric effect equation
1) Instead of giving you kinetic energy, the problem will give you velocity. From there you have to calculate kinetic energy using equation: KE = 1/2mv^2 where m=mass of an electron in kilograms (kg) 2) Instead of giving you the work function in Joules like the equation needs, they'll give you a value in electron volts (eV). In this situation, you simply convert from eV to J knowing that 1 eV = 1.60 x 10^-19 Joules 3) Instead of giving you frequency (v) you'll be given wavelength (λ). So, you will have to use this equation to find the frequency: v = c/λ where c=speed of light and λ=wavelength 4) Instead of being given h or v, you'll be given total energy. In this scenario, you can just plug in energy for hv in the equation KEe = hv - ϕ, because E=hv
1) If the energy of the light aka E = hv is less than the work function, are electrons are ejected (hv < ϕ) ? 2) If the energy of the light aka E = hv is greater than the work function, are electrons are ejected (hv > ϕ)? 3) The greater the light energy is than the work function (ϕ), the more ____ BECAUSE ___
1) No, there are no electrons are ejected when hv < ϕ 2) Electrons are ejected when hv > ϕ 3) The greater the light energy is, the more kinetic energy the electron is going to have BECAUSE a certain amount of energy is required to remove an electron all of the leftover energy becomes kinetic energy in the electron
Photoelectric effect key points
1) Shining light onto a metal surface in a vacuum can eject electrons from the metal surface 2) The light must have a minimum frequency to eject electrons from the metal surface 3) The minimum frequency is different for different metals 4) Increasing the brightness of the light ejects more electrons, but the electrons have the same kinetic energy as before 5) Increasing the frequency of light ejects the same number of electrons but with higher kinetic energy
Energy units: joules vs. electron volts
1) Work function is energy required to remove one electron from the metal. No matter which metal you use, this is a very small amount of energy. 2) typical ϕ values range form 3 x 10^-19 to 6 x 10^-19 J (Joules) 3) ϕ is often given in electron volts eV instead 4) 1 eV = 1.6022 x 10^-19 J 5) Typical ϕ values range from 2.5 eV to 5.5. eV
Define work function and what is its symbol
1) amount of energy required to release an electron 2) represented by phi symbol ϕ
If light is a wave, then (3)
1) increasing the brightness (intensity) increased the energy of the light. (ACTUALLY IDK ABOUT THIS?? INCREASING INTENSITY ACTUALLY DOES INCREASE THE AMOUNT OF ELECTRONS, IT DOES NOT INCREASE THE ENERGY OR THE KINETIC ENERGY.) This means you would get the same number of electrons ejected, but the ejected electrons would have a higher kinetic energy. (ACTUALLY, WHEN YOU CHANGE/INCREASE THE ENERGY (IN JOULES OR ELECTRON VOLTS eV, YOU INCREASE THE KINETIC ENERGY, BUT THE NUMBER OF ELECTRONS (e^-) STAY THE SAME) 2) decreasing the brightness would result in the same number of electrons being ejected, but with a lower kinetic energy. 3) a wave of light falls over an area of the metal surface, spreading its energy across a large number of electrons. This means that at first there would be no electrons ejected, but after the light has been applied to the surface for a short time, electrons should absorb enough energy to start being ejected from the entire area that was struck by the light.
Photons (4)
1) they're particles of light 2) light exists as tiny packets of energy (later called photons) 3) energy of light is proportional to its frequency-blue light has greater energy than red light at the same intensity 4) energy of a photon of light is equal to Planck's constant times the frequency of the light (E = hv)
Planck's constant
6.626 x 10^-34 J x s, where J is the symbol for the joule (Joules per second)
When an electron is ejected from the metal surface, where does thee energy of the photon go?
A & B: A. overcoming potential energy that holds an electron to the metal B. Into the kinetic energy of the electron (see photo)
Consider a metal surface with a work function of 5 eV. For each scenario describe the effect on (1) the number of electrons ejected from the metal surface and (2) the KE of those electrons A. You shine light with energy of 2 eV on the metal surface. B. You increase the intensity of the 2 eV light. C. You change the energy of the light to 8 eV. D. You change the energy of the light to 11 eV. E. You increase the intensity of the 11 eV light.
A) Energy of the photon (remember E = hv) = 2 eV Work function = 5 eV KEe = hv - ϕ KEe = 2 eV - 5 eV (1) The energy of the photon is less than the work function so you won' t be able to get an electron at all! (2) There is no KE bc there are no electrons B) (1) Even though you INCREASE THE ENERGY of the 2 eV light, it's still only 2 eV and therefore still less than the work function. There still isn' t enough energy bc the work function is greater than the energy of the photon. (2) There is no KE bc there are no electrons C) Energy of the photon (remember E = hv) = 8 eV Work function = 5 eV KEe = hv - ϕ KEe = 8 eV - 5 eV KEe = 3 eV (1) The energy of the photon is greater than the work function so you now have electrons! (2) The kinetic energy of the electrons is 3 eV or to find it in Joules multiply 3 eV x (1 x 10^-19 J) D) Energy of the photon (remember E = hv) = 11 eV Work function = 5 eV KEe = hv - ϕ KEe = 11 eV - 5 eV KEe = 6 eV (1) The energy of the photon is greater than the work function so you have electrons! However, you still have the same number of electrons as in part C. This is because of this fundamental concept from "If light is a wave": Increasing the energy of the light (not increasing the intensity) gets you the same number of electrons ejected, but the ejected electrons would have a higher kinetic energy. Aka, just increasing the energy of the electrons WON'T give you more electrons, but it will increase their kinetic energy. (2) The kinetic energy of the electrons is 6 eV or to find it in Joules multiply 6 eV x (1 x 10^-19 J). This value is twice the amount of kinetic energy that the electrons had in part C! D) (1) Because you increased the intensity/energy of the 11 eV light, you will eject more electrons. (2) The kinetic energy is NOT affected when you increase the intensity of the eV light, it's only affected when you increase the energy.
As the wavelength goes down aka as it ____, the frequency ____ AND the energy ____
As the wavelength goes down aka as it gets shorter, the frequency goes up (higher frequency) AND the energy goes up (higher energy)
Each metal has a minimum ____ and a maximum ____ + (see graph)
Each metal has a minimum threshold frequency and a maximum wavelength (REMEMBER WAVELENGTH AND FREQUENCY ARE INVERSELY RELATED) + (see graph)
What is the longest wavelength of light (lowest energy) capable of ejecting an electron from potassium in a vacuum? Which portion of the electromagnetic spectrum does this photon fall in? φ = 3.67 × 10-19 J h = 6.626 × 10-34 J·s c = 3.0 × 108 m/s
Final answer is 542 nm which is in visible region of light which is from 400 nm to 700 nm and in the green light range (see photo)
Example of graphs (may have to do this on test)
Graphs Explained: A) You start with getting no electrons for a long time, then you get electrons but that number of electrons stays the same B) At first you start with no kinetic energy bc there's no electrons. Then as the energy increases, the kinetic energy of the electrons increases. C) As the intensity (of the brightness) increases, we get more electrons D) As the intensity (of the brightness) increases, the kinetic energy stays the same
Einstein's explanation for why increasing the brightness of the light ejects more electrons, but the electrons have the same kinetic energy as before
His explanation: 1) light is acting like tiny particles or wave packets. Each packet of light ejects one electron. 2) Energy of light particle is proportional to its frequency. Higher frequency means light particle has more energy. 3) Increasing brightness of light means more light particles are striking the metal
Heinrich Hertz
In 1887 Heinrich Hertz discovered that electricity can arc across a longer distance (such as a gap in a wire) when ultraviolet light is shined onto the wire. This phenomenon became known as the photoelectric effect. When scientists tried to explain how light waves made this happen, they ran into some major problems.
What determines the energy of the wave?
In the wave theory of light, different colors of light do not have different energies. It is the brightness of the light, not the wavelength or frequency, that determines the energy of the wave.
Read
In wave theory, a higher amplitude wave has greater energy. For light, a higher amplitude corresponds to a brighter light. Therefore, according to the wave theory of light, a brighter light should have greater energy than a dim light at the same wavelength, and should therefore eject electrons with higher kinetic energy.
What is the unit of work function?
Joules (J)
Formula for kinetic energy of an electron (Photoelectric effect)
KEelectron = hv - ϕ (see photo) AND KEEP IN MIND KE = 1/2mv^2 if velocity is ever involved
What effect did you see when you increased the brightness of the light in Step 3 of the experiment?
More electrons were ejected, but the ejected electrons had the same kinetic energy as the electrons that were ejected using dimmer light.
Visible light range scale in nanometers?
Visible light has a narrow range of wavelengths from about 400 nm (violet light) to about 700 nm (red light).
When frequency goes up, energy ____. When frequency goes down, energy ____.
When frequency goes up, energy goes up. When frequency goes down, energy goes down.
When the problem asks for the longest wavelength, lowest frequency, or lowest energy capable e of ejecting an electron from potassium in a vacuum, _____
plug in 0 (zero) for KE in the formula KEe = hv - ϕ
When the wavelength of the light is shorter and the frequency is higher...
the kinetic energy of the ejected electrons will be higher
