Post on 07-Dec-2015
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1 Table of Values
Mass of One Electron: 9.1× 10−31
Charge on the electron: q = 1.6× 10−19
Planck’s Constant:h = 6.63× 10−34Jsh = 4.14× 10−15eV s
Electronvolt 1 eV = 1.6× 10−19 J
2 Formulae
Photoelectric Effect:Ek(max) = hf −W = qV0
= hf − hf0Photon Energy E = hf
de Broglie Wavelength: λ =h
p=
h
mvv, f and λ: v = fλ v = fλ
Bohr Hydrogen E-Levels En =−13.6
n2
W - Work Function
V0 - Stopping Potential (V)
f0 - Threshold Frequency
3 Properties
Property Reflection Refraction DiffractionWave: 3 3 3
Particle: 3 7 7
Property Interference Photoelectric EffectWave: 3 7
Particle: 7 3
3.1 De Broglie’s Wavelength
Single Slits
Red has a higher λ then blue
Double Slits
Path Difference:(bright) PD = 0, λ, 2λ, . . .(dark)PD = λ
2 ,32 ,
52 , . . .
3.2 Young’s Double Slit Experiment
W =λL
dW - fringe spacing
λ - Wavelength
L - slit to screen distance
d - distance between slits
4 Examples
4.1 Question 1
Cesium metal is illuminated by green light with a wavelengthof 550mm.
1. Calculate the energy of a photon green lighth = 4.14× 10−15eV , c = 3.00× 108ms−1)
E = hf
→ E =hc
f
→ E =4.14× 10−15 × 3.0× 108
550× 10−9
Ek = 2.26 eV
The work function of cesium is 2.10 eV.
2. Calculate the maximum kinetic energy of the electronsejected from the metal surface when green light illumi-nates cesium metals.
Ek(max) = hf −WEk(max) = Eph −WEk(max) = 2.26− 2.10
Ek(max) = 0.16 eV
1
LIGHT C
HEAT SHEET
VCE PHYSICS 20
15UNIT 4
AOS 2
by Eric
JIANG
VCE Physics Unit 4 AOS 2 - Light
Copyright (C) 2015 by Eric JIANG
Violet light now illuminates cesium metal and the maxi-mum kinetic energy of the photoelectrons is 2.80 eV.
3. Show that the maximum speed of the electrons ejectedfrom the metal surface is 9.9× 105 ms−1.
Ek(max) = 2.8 eV = 2.8× 1.6× 10−19
Ek(max) = 4.48× 10−19 J12mv
2 = 4.48× 10−19
v2 =2× 4.48× 10−19
9.1× 10−31
v =
√2× 4.48× 10−19
9.1× 10−31
v = 9.9× 105 ms−1
4.2 Question 2
An electron gun accelerates electrons across a potential dif-ference of 2500 V. The initial speed of the electrons can beconsidered to be almost zero.
Show that the final speed of the electrons is approximately10% of the speed of light.
12mv
2 = eV
v =
√1.6× 10−19× 2500× 2
9.1× 10−31
= 2.96× 107 ms−1
c = 3.0× 108
c× 10% = 3.0× 108 × 0.1c× 10% = 3.0× 107
2.96× 107 ms−1 ≈ 3.0× 107
⇒ v ≈ 10%× c
In an experiment to demonstrate the photoelectric effect,physics students allow lights of various frequencies to fallon a metal surface in a photocell. The photoelectrons aredecelerated across a retarding voltage, and the stoppingpotential, Vs, is measured for each frequency. The data theyobtained is graphed in Figure 1.
Figure 1
The students use the data points on the graph to determinethe value for the work function of the metal.
1. Determine the magnitude and the unit of the workfunction for this metal surface.W = qV
W = 1.6× 10−19 ×−2W = −3.2× 10−19 JMagnitude: 3.2 Unit: J-or-Magnitude: -2 Unit: eV
2. What is the maximum kinetic energy (in eV) of the pho-toelectrons produced when ultraviolet light of frequency1.93× 1016 Hz is incident on the metal surface?
Ek(max) = hf −WEk(max) = 4.14× 10−15 × 1.93× 1016 − 2
Ek(max) = 79.902− 2Ek(max) = 77.902
Ek(max) = 77.9 eV
4.3 Question 3
A radiolarian is taken with a electron microscope and a opticalmicroscope. The electron microscope gives a clearer imagethan the optical microscope.
1. Explain why the electron microscope gives a clearer im-age than the optical microscope.
Visble light has a longer wavelength than the electronsand will exhibit more diffraction, giving a more fuzzierimage compared to the electron microscope
2
LIGHT C
HEAT SHEET
VCE PHYSICS 20
15 U
NIT 4 AOS 2
by Eric
JIANG
VCE Physics Unit 4 AOS 2 - Light
Copyright (C) 2015 by Eric JIANG