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Photoelectric effect
p. 275
The photoelectric effect is when light shines on metal and electrons are released from the metal.
Valensband
Conduction band
-
Valence band-
Effect of intensity
Valensband
Conduction band
-
Valence band-
-
-
-
-
Threshold frequencyThe minimum frequency of light required to release electrons from a certain metal surface. (f0)
Work functionThe working function of a metal is the minimum energy required by an electron in the metal to be released from the metal surface. (W0)
PhotonsLight is emitted as elementary particles (quantum) and is called photons.
Energy of photons
Eαf
E = hf = ℎ𝑐
𝜆
Planck's constant: h = 6,63 x 10-34 J.s
Energie
E = hf
e-
e-
E = W0 + Ek
Energy E = W0 + Ek
hf = hf0 + ½mv2
Kinetic energy of photoelectrons released is determined by the frequency of the light.
or ℎ𝑐
𝜆= hf0 + ½mv2
What is the meaning of the photoelectric effect?
• Establish the quantum theory• Demonstrate the particle of light.
0 1
A
e ee
e
ee
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e
e
e
e
e
e
e
e
ee
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Wo = hfo
E = hf
E =ℎ𝑐
𝜆
Ek = 1
2𝑚v2
E = Wo + Ek
0 1
A
e ee
e
ee
e
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Low fLong λ
0 1
Ae
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Low f
High intensity
e ee
ee
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0 1
eAHigh fshort λ
e ee
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0 1
e
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A
High intensity
High f
e ee
ee
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Homework
Exercise 1 p 3242, 5, 7, 12, 22
Emission and absorption spectra
p. ...
Spectra
Light ray is refracted to demonstrate the frequencies of all the colours in a continuous spectrum.
Diffraction gratingSpectroscope
Spectra
Spectra
Emission spectra Absorption spectra
Continuous
Veroorsaak deur ligbron
Line emission
Spectra
Emission spectrum of gasses - Hydrogen
n=1
n=2
n=3
n=4
n=5
n=6
ground state
1st excited state
--
--
Infrared series
Ultravioletseries
Internal energy of an atom
n=1 (E1)
n=2 (E2)
n=3 (E3)
Energy levels
-
When internal energy is lost, the energy is emitted as a photon:E = E3 – E2 = hf
Example: For the
electron crossing of energy levels E3
to E2 calculate the energy of an emitted photon.
n=1 (E1)
n=2 (E2)
n=3 (E3)
-1,2 x 10-19 J
-
E = E3 – E2 = hf= -2,614 x 10-19 - (-5,424 x 10-19)= 3,81 x 10-19 J
-2,614 x 10-19 J
-5,424 x 10-19 J
-21,76 x 10-19 J
Example: For the
electron crossing of energy levels E3
to E2 calculate the frequency of an emitted photon.
n=1 (E1)
n=2 (E2)
n=3 (E3)
-1,2 x 10-19 J
-
E = hf3,81 x 10-19 = 6,63 x 10-19 x ff = 4,24 x 1014 Hz
-2,614 x 10-19 J
-5,424 x 10-19 J
-21,76 x 10-19 J
Example: It is found that the wavelength of the indigo line in the hydrogen spectrum is
approximately 434 nm. Calculate the energy of a quantum (photon) of the indigo light.
E = ℎ𝑐
𝜆
= (6,63×10−34)(3×108)
4,34×10−7
= 4,58 x 10-19 J
Absorption spectra
When light passes through a cold gas light (photons) is absorbed so that electrons move to a higher energy level. The light is emitted again (in other directions). The light that passes through the gas shows a spectrum without the absorbed light.
Absorpsiespektrum Emisiesiespektrum
Huiswerk
P 3422, 6, 9