Photoelectric Effect -01- Work Function

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The photoelectric threshold of a certain metal is 2750 Å. Find the work function of electron emission from this metal.  Watch videos related to MODERN PHYSICS: https://dailymotion.com/playlist/x90qrm

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00:00Hi friends, start learning now, and later you'll be grateful you did.

00:09Albert Einstein won the Nobel Prize for his research on the photoelectric effect.

00:15One of the terms in the photoelectric effect is the work function.

00:19We will learn to calculate the value of the work function.

00:25Before we go any further, the problem states the threshold wavelength.

00:30Lambda Th is equal to 3200 angstroms.

00:35Th is short for threshold.

00:41You may not know that 1 angstrom is equal to 10 to the power of minus 10 meters.

00:48There's the Planck constant, and the speed of light in a vacuum.

00:55As an illustration, here's a metal surface.

01:00On the metal surface are several atoms.

01:04As usual, these atoms consist of atomic nuclei and electrons orbiting them.

01:09To simplify the animation, we'll use a snapshot of an electron at a specific moment.

01:17When a beam of light is directed toward the surface,

01:24there's a chance that this photon will be absorbed by the electron.

01:29This electron will be released from the electron cloud and move toward the metal surface.

01:37This electron will remain stationary near the surface.

01:40The energy absorbed by the electron to escape its atomic bonds is known as the work function.

01:49What's important to understand is that the work function only causes electrons to be released from atomic bonds.

01:55The electrons themselves are still immobile.

01:57Mathematically, the work function can be calculated using the equation phi equals hc per lambda th.

02:06It seems the values of these three physical quantities are already known.

02:16If the numbers are like this, we'll need to use a scientific calculator.

02:20The result is approximately 6.2119 times 10 to the power of minus 19 jowls.

02:31In electron volts, simply divide this value by 1.6 times 10 to the power of minus 19.

02:39Phi is approximately 3.88 electron volts.

02:42A tip for those of you who encounter wavelengths in angstroms, just use this formula.

02:54Lambda th here is 3,200, because this value must be in angstroms.

03:01The results are also similar.

03:05For those of you who are quick to memorize, just use this formula.

03:09Happy learning, everyone.

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