2
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 -34 Js h =4.14 × 10 -15 eV s Electronvolt 1 eV =1.6 × 10 -19 J 2 Formulae Photoelectric Eﬀect: E k(max) = hf - W = qV 0 = hf - hf 0 Photon Energy E = hf de Broglie Wavelength: λ = h p = h mv v, f and λ: v = v = Bohr Hydrogen E-Levels E n = -13.6 n 2 W - Work Function V 0 - Stopping Potential (V) f 0 - Threshold Frequency 3 Properties Property Reﬂection Refraction Diﬀraction Wave: Particle: Property Interference Photoelectric Eﬀect Wave: Particle: 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 , 3 2 , 5 2 ,... 3.2 Young’s Double Slit Experiment W = λL d W - 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 wavelength of 550mm. 1. Calculate the energy of a photon green light h =4.14 × 10 -15 eV , c =3.00 × 10 8 ms -1 ) E = hf E = hc f E = 4.14 × 10 -15 × 3.0 × 10 8 550 × 10 -9 E k = 2.26 eV The work function of cesium is 2.10 eV. 2. Calculate the maximum kinetic energy of the electrons ejected from the metal surface when green light illumi- nates cesium metals. E k(max) = hf - W E k(max) = E ph - W E k(max) =2.26 - 2.10 E k(max) =0.16 eV 1 LIGHT CHEAT SHEET VCE PHYSICS 2015 UNIT 4 AOS 2 by Eric JIANG VCE Physics Unit 4 AOS 2 - Light Copyright (C) 2015 by Eric JIANG

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lIGHT cHEAT SHEET

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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