Transcript
58
Marks Total
ASSIGNMENT 13
Part One: Blackbody Radiation and the Quantum
Part One of this assignment is worth 27 marks. The value of each question is noted in parentheses in the left margin. Note: The answer areas will expand to fit the length of your response.
(3) 1.
Describe the pattern of intensity vs wavelength observed in blackbody experiments and explain why these observations were in conflict with the prediction of classical physics.
Answer:
Classical physics suggested that more energy would be produced at higher frequencies. The shorter the wavelength, the greater the intensity.
(2) 2.
What is the frequency of a 13.0 eV photon?
Answer:
E = 13 x 1.6 x 10-19 J
E = 20.8 x 10-19 J
E = hc/?
E = h(?f)/? = hf
f = E/h
E = Energy of Photon in Joules
h = Plank's constant = 6.626 x10-34.
? = wave length
c = Speed of light = 3 x108 m/s.
f = frequency of a photon in Hz
f = E/h
Substitute : E = 20.8 x 10-19 , h = 6.626 x10-34
f = (20.8 x 10-19)/( 6.626 x10-34)
f = 3.13915 x 1015 Hz
(2) 3.
What is the wavelength of a 4.30 x 10-19 J photon?
Answer:
(6.626x10-34)(2.998*108) /(4.36*10-19) = 4.55613*107
(4) 4.
How many photons with a wavelength of 532 nm are released each second by a 5.0 mW laser?
Answer:
Energy for each photon = E = hf = hc/?
E = (6.63 x 10-34 J-s)(3 x 108 m/s)/(532 x 10-9 m) = 3.740 x 10-19 J/photon
P = NE per sec, so in each second, there are N photons emitting
N = P/E = (5.0 x 10-3 J/s)(3.143 x 10-19 J/photon)
= 1.87 x 10-21 photons/sec
(2) 5.
When welders are working on pipelines they wear filters that block UV light. Why is this safety precaution necessary?
Answer:
To protect from premature aging of the skin, suppression of the immune system, damage to the eyes, and skin cancer and other dangerous diseases and such.
(3) 6.
A person lives near a 50 kW radio tower. Why is it safe to live near a high power radio tower but dangerous to be exposed to an x-ray machine that uses 7.5 kW?
Answer:
First the radio tower disperses much more, for instance the person in the first example probably lives a kilometer or more from the tower while you are a meter or two from the x-ray machine when it becomes dangerous. Moreover, the photons from x-ray are of much greater energy, and the energy of each one can be more important then the total number of photons.
For instance, radio waves carry less energy per photon than visible light, while X-rays carries more energy per photon than visible light.
(2) 7.
Why did the introduction of the photon effectively end classical physics?
Answer:
It was the first "particle" observed to have both the properties of a true particle, and the properties of waves. Thomas Young's double-slit experiment proved that photons behave like electromagnetic waves when unobserved, and like particles while under observation.
(2) 8.
How did Einstein’s explanation of the photon contradict the classical explanation of light?
Answer:
Einstein modified Planck's concept of quantized energy to describe the experimental results. Einstein proposed that light could travel in small quantized packets of energy (photons) instead of strictly behaving as a classical wave. Einstein showed that the kinetic energy of the ejected electrons was equal to the energy of the incident photon (h?) minus the energy barrier to releasing an electron from that particular metal (work function = ?). This interpretation is described by the following equation:
KE = 1/2 mv2 = h???
This model was able to fully account for the experimental results including the lack of dependence of the energy of the ejected photons on the intensity of the incident radiation as well as the failure of some frequencies of light to eject any photons (the incident energy of the photon was less than the work function).
(1) 9.
Provide an example of an experiment that Einstein’s theory could not explain.
Answer:
Gravitational Waves
10.
(1)
Using the concepts of quantum and photons, explain the difference between a 100 W and a 40 W light bulb if they both produce only green light. Answer the following questions in your explanation.
Why does the 100 W bulb appear brighter than a 40 W bulb?
Answer:
The 100W bulb appears brighter because it produces a greater number of photons every second.
(1)
Why would you expect the 100 W bulb to feel hotter than the 40 W bulb?
Answer:
The intensity then the energy flow of 100 W is > than 40 W light bulb
The absorbed energy transformed in heat of 100 W light bulb is > than 40 W light bulb
(3)
If 10% of the energy consumed by each bulb produces green light of 500 nm, how many green coloured photons does each bulb produce each second?
Answer:
W = n h c / ?
n1 = 100 ? / (h c) = 2.5x10^20 photons/second [100 W]
n2 = 40 ? / (h c) = 1x10^20 photons/second [40 W]
(1)
How does your answer to 10c support your explanation of brightness?
Answer:
IT WAS THE FIRST "PARTICLE" OBSERVED TO HAVE BOTH THE PROPERTIES OFA TRUE PARTICLE, AND THE PROPERTIES OF WAVES. THOMAS YOUNG'SDOUBLE-SLIT EXPERIMENT PROVED THAT PHOTONS BEHAVE LIKEELECTROMAGNETIC WAVES WHEN UNOBSERVED, AND LIKE PARTICLES WHILEUNDER OBSERVATION.
STOP!
When you have completed all of the questions in Part One, save your work to your desktop. You will return to this assignment to complete Part Two after you have completed the remainder of the content in the next section.
Part Two: The Photoelectric Effect
Part Two of this assignment is worth 31 marks. The value of each question is noted in the left margin in parenthesis. Note: The answer areas will expand to fit the length of your response.
1.
(3)
While conducting research into the design of a light sensor, scientists measured the kinetic energy of photoelectrons that had been ejected from an unknown metal surface. The metal was exposed to various frequencies of EMR and the stopping voltage was measured to obtain the kinetic energy of the photoelectrons. The following data was obtained.
Incident EMR frequency
x 1014 Hz
Kinetic energy of photoelectrons (eV)
6.0
0.38
7.0
0.80
8.0
1.20
9.0
1.63
10.0
2.04
Using the data above, plot a kinetic energy vs. frequency graph and use it to answer questions b – e.
Answer:
(2)
According to your graph, what is the threshold frequency of the unknown metal? Label the threshold frequency on your graph.
Answer:
(2)
Determine the work function of the unknown metal.
Answer:
Hello =] I believe I just answered this question for my homework. This is what I did:
Work function = hc/wavelength - KE.
So you get ((6.63*10^ -34 Joules/s)(3*10^8 m/s))/(162*10^ -9 m) - 5.46*10^-19 J.
As you can see, the seconds cancel out in the numerator and meters cancel out between the numerator and denominator. You have to remember that since the constant "c" is the speed of light and it has the units meters per second, where the wavelength is in nm. So you have to multiply your wavelength by 10^-9 to convert it into meters.
(1)
Using the tables below, identify the unknown metal.
Element
Aluminum
Beryllium
Cadmium
Calcium
Carbon
Cesium
Copper
Work Function
(eV)
4.08
5.00
4.07
2.90
4.81
2.10
4.70
Element
Magnesium
Mercury
Potassium
Selenium
Sodium
Zinc
Work Function
(eV)
3.68
4.50
2.30
5.11
2.28
4.33
Answer:
Alluminium.
(3)
Explain how to find Planck’s constant from your graph then, using your graph, determine the experimental value for Planck’s constant in eVs.
Answer:
what I did:
Work function = hc/wavelength - KE.
So you get ((6.63*10^ -34 Joules/s)(3*10^8 m/s))/(162*10^ -9 m) - 5.46*10^-19 J.
As you can see, the seconds cancel out in the numerator and meters cancel out between the numerator and denominator. You have to remember that since the constant "c" is the speed of light and it has the units meters per second, where the wavelength is in nm. So you have to multiply your wavelength by 10^-9 to convert it into meters.
(2) 2.
What is the energy of a photon that has a wavelength of 460 nm?
Answer:
F = C /wavelength
f = 300,000,000 /(4.6 x 10-7)
f = 6.52173913 × 1014 Hz
E= hf
E= (6.62 X 10-34) x (6.52173913 × 1014)
E= 4.3173913 × 10-19 J
(2) 3.
A photoelectric surface has a work function of 2.00 eV. What is the threshold frequency of this surface?
Answer:
2.00 eV = 3.2044e-19 J
Work function w = h * f0 .
f0 = w / h
f0 = ( 3.20x10-19 ) / (6.63 × 10-34 )
= 4.83 x 1014 Hz
(3) 4.
A photon of frequency 8.2 x 1015 Hz is incident upon a photoelectric apparatus containing a metal whose threshold frequency is 3.6 x 1015 Hz. What is the stopping voltage?
Answer:
Ek is kinetic energy .
q is the charge of electron (1.6 × 10–19 C )
The kinetic energy is Ek = Ep - w ? hf - hf0 ? h(f -f0 )
Ek = h(f -f0 )
Where is plank's constant = 6.63 × 10-34 J·sec .
Ek = 6.63 × 10-34 ( 8.2 - 3.6 )x 1015 .
Ek = 6.63 × 4.6 × 10-19 ? 30.498 × 10-19 eV.
Vstop = Ek / q ? (30.498 × 10-19 ) / (1.6 × 10–19 )
= 19.06 V
(3) 5.
Electrons are ejected from a photoelectric cell with a maximum kinetic energy of 1.20 eV. If the incident light has a wavelength of 410 nm, what is the work function of the cell?
Answer:
K_m = hc/? - ?
? = hc/? - K_m = 2.92573068902439e-19 J
? = 2.93E-19 J
(4) 6.
Light with a wavelength of 425 nm falls on a photoelectric surface that has a work function of 2.00 eV. What is the maximum speed of any emitted photoelectrons?
Answer:
Wave length of light ?= 425 nm = 425 x 10-9 m .
Work function w = 2 eV = 3.2043 x 10-19 J .
The kinetic energy of light Ek = Ep - w ? [ (hc)/?] - w
Where h is plank's constant = 6.63 × 10-34 J sec
c = velocity of light = 3 × 108 m/s
Ek = [ (6.63 × 10-34 × 3 × 108) / 425 x 10-9 ] - 3.2043 x 10-19 J
Ek = [ 4.6772 x 10-19] - 3.2043 x 10-19 J
Ek = 1.4729 x 10-19 J .
The kinetic energy of light Ek = (1/2) mv²
v = ? [ (2Ek)/m]
Where m is mass of electron 9.11 x 10-31 kg.
v = ? [ (2 x1.4729 x 10-19) / 9.11 x 10-31]
v = 5.686 x 105 m/s .
(6) 7.
Explain the photoelectric effect in terms of energy. In your explanation, start with the energy of the incident EMR photons and finish with the energy of the stopping voltage. Be sure to include the following terms in your answer: threshold frequency, light intensity, photoelectron current, work function, photoelectron kinetic energy and stopping voltage.
Answer:
The basic notion of the photoelectric effect: A photon hits an atom and knocks an electron out.
Knocking the electron out of the atom uses up a certain amount of the energy in the photon (= the work function), and the rest of the energy shows up as kinetic energy in the photoelectron. (This is the *photoelectron kinetic energy*.)
The energy of a photon is proportional to the frequency the wave. In particular,
eq. 1) E = hf.
The amount of energy that is just enough to knock an electron loose is the *work function*, and the energy corresponds to the *threshold frequency*, thanks to equation (1).
As the energy (frequency) of the incoming photon is increased, more energy is available to go in to moving the electron, or photoelectron kinetic energy. As the kinetic energy increases, the electron is not only moving faster, but has enough energy to carry it uphill against an electrical potential (voltage).
Because the energy of light depends only on the wave frequency, increasing the intensity won't increase the energy in the ejected photoelectrons. What it will do is, because the number of photons incident on the photocell increases, the number of photoelectrons ejected also increases. Star light will produce a very, very weak current in a photocell, with the same voltage as sunlight. But sunlight will produce a much stronger current.
To measure this energy, one puts an electrode near a source of photoelectrons, and arranges the electric field so that electrons have to cross a voltage gap in order to reach that electrode. When the voltage reaches a critical voltage for any given frequency of light, no more electrons make it uphill to the electrode. The current flow stops. This is the "stopping voltage".
If the stopping voltage is measured in volts, then since the charge of each electron is one electron's worth of charge, the kinetic energy of the photoelectron is equal to the stopping voltage, but the unit is electron volts.
When you have completed all of the questions in this assignment, submit your work.
