Gen Chem_Ch 6_9th_Photoelectric Effect (2)
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Overview
Dr. Reza Latifi applies the photoelectric equation to 400 nm light, converting wavelength to 4.00 × 10⁻⁷ m and calculating photon energy as 4.97 × 10⁻¹⁹ J. The worked calculation uses a 2 eV work function and obtains a maximum electron kinetic energy of about 1.76 × 10⁻¹⁹ J, although the problem is initially stated with a 1.2 eV work function; with 1.2 eV, the kinetic energy would instead be about 3.05 × 10⁻¹⁹ J.
Key takeaways
- For 400 nm light, E = hc/λ gives a photon energy of approximately 4.97 × 10⁻¹⁹ J, or 3.10 eV.
- The photoelectric equation is Kmax = E_photon − φ; electron emission occurs when photon energy exceeds the work function.
- The worked calculation uses φ = 2 eV, producing Kmax ≈ 1.76 × 10⁻¹⁹ J, but the problem is initially described as having φ = 1.2 eV.
- With the stated 1.2 eV work function, the same 400 nm photons would produce Kmax ≈ 3.05 × 10⁻¹⁹ J.
- Doubling intensity at fixed frequency increases the number of emitted electrons and current, but does not change their maximum kinetic energy.
Chapters
- The problem asks whether 400 nm light ejects electrons, what their maximum kinetic energy is, and what doubling intensity changes.
- Latifi initially states the metal work function as 1.2 eV and notes that photon energy and work function must use matching units.
- Convert 400 nm to 4.00 × 10⁻⁷ m for use in the photon-energy equation.
- Using E = hc/λ, with h = 6.626 × 10⁻³⁴ J·s and c = 3.00 × 10⁸ m/s, gives photon energy of 4.97 × 10⁻¹⁹ J.
- The calculation converts 2 eV to 3.20 × 10⁻¹⁹ J, despite the problem initially specifying a 1.2 eV work function.
- Subtracting the 2 eV work function yields a maximum kinetic energy of 1.76 × 10⁻¹⁹ J, so electrons are emitted.
- Using the initially stated 1.2 eV work function instead would give a maximum kinetic energy of approximately 3.05 × 10⁻¹⁹ J.
- At unchanged wavelength and frequency, photon energy remains constant, so each emitted electron has the same maximum kinetic energy.
- Doubling intensity means more photons strike the metal surface, resulting in more emitted electrons and increased current.
- Intensity affects the number of emitted electrons, while frequency determines their maximum kinetic energy.
Transcript excerpt
0:11 [music] >> Hello and welcome back and thank you very much for watching another lecture from general chemistry. I want to practice one more question about photoelectric effect. They gave me a metal that they say work function for it or binding energy is 1.2 electron volt and then the light that we are sending it has wavelength equal to 400 nanometer. First the question it says are we going to have any electron ejected from the system? Calculate the maximum kinetic energy of the emitted electron and what's going to happen if we you know let's say it says what happened to the number of maximum kinetic energy of emitted electrons if we if the light intensity doubled? Light
0:59 intensity, okay? Not the wavelength or frequency. So, highlight that one and we're going to come after it. Okay, light intensity, not the wavelength or frequency. And so, in order to answer all this question, we need to know what's the binding energy in joule and also how what the energy that we are applying in joule or you can you know use both of them in electron volt. It doesn't matter and but in any case you need to have the same units. So, I go the binding energy or fee Okay. It is given as equal to two electron volt and one electron volt is equal to 1.6
1:47 02 * 10 to the power of minus 19 joule. And I'm going to remind you that the energy the total energy that we are applying is equal to fee plus the kinetic energy. Or you can say the here the kinetic energy or just simply say K. And uh they give us the wavelength for the emitted for light wavelength for the light that we are applying and uh
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Related notes
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Dr. Reza Latifi at Oklahoma State University, OSU.