The Photoelectric Effect: Difference between revisions
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===Middling=== | ===Middling=== | ||
Question: An unknown material has a work function value Φ = 2.29 eV and ejects a photoelectron at | Question: An unknown material has a work function value Φ = 2.29 eV and ejects a photoelectron at 8900 m/s. What is the energy of the photon that struck that material? | ||
Answer: The work | Answer: The energy relation for the photoelectric effect is given as <math>{E_f = K+Φ}</math>. | ||
The work function value has been given but is in eV. You can give your answer in eV but either way, you will need to do a conversion. The relation between eV and Joules is <math>{1 eV = 1.602 * 10^-19 J}</math>. | |||
You need to calculate the kinetic energy of the photoelectron. At a velocity of 8900 m/s and electron mass of <math>{9.11*10^-31}. Using <math>{K = 0.5mv^2}</math>, we have that the kinetic energy is <math>{3.61 * 10^-23 J}</math>. | |||
Combining with the work function, we have <math>{3.61 * 10^-23 J + (1.602 * 10^-19)(2.29) = E_k}</math>. | |||
We then have that <math>{E_k = 3.66 * 10^-19 J}</math>. | |||
===More Difficult=== | |||
Question: A beam of electrons is shot at a metal plate with a frequency of 400 Hz and an electron is observed as being emitted with 1.65*10^31 J of energy. Another beam is shot at the plate at 300 Hz. Will the photoelectric effect be observed? | |||
Answer: The problem first requires that the work function for the metal be calculated. The work function is defined as <math>{E_f - k = Φ}</math>. | |||
The energy must be calculated from the frequency using the relation <math>{hf = E_f}</math>. Using this, you should get <math>{ E_f = 2.65*10^-31 J}</math>. | |||
Using this, you should find that <math>{Φ = 1 * 10^-31 J}</math>. | |||
Now applying this to the second beam of electrons and calculating the energy from the given frequency, you should have <math>{1.988*10^-31 J - 1 *10^-31 J = K}</math>. You can see that the kinetic energy will be positive, and thus may conclude that the photoelectric effect will be observed. | |||
== See also == | == See also == |
Latest revision as of 23:47, 24 April 2022
Claimed by Joe Baldino 4/16/2022 Short Description of Topic
The Main Idea
The photoelectric effect is the phenomena in which electrons are emitted from a material that is bombarded by electromagnetic radiation. First observed in the 19th century, the effect was confounding to scientists because of its violation of classical electromagnetism. These discrepancies ultimately led to Albert Einstein making groundbreaking proposals about the nature of light.
History
German physicist Heinrich Hertz is credited with the discovery of the photoelectric effect in 1887 when he observed a changing of sparking voltage between electrodes when ultraviolet light is shined on them[1]. The effect was subsequently studied by various other notable physicists, including Aleksandr Stoletov and J.J. Thomson. Most significant of this period, however, were the studies undertaken by Philipp Lenard. Lenard extensively worked on researching the photoelectric effect and determined that the velocity at which electrons are emitted from a material is independent of the intensity of the light[2]. This was one of the major discoveries that directly violated what was though to be known about electromagnetic radiation. This, compounded with later studies showing that there is a threshold frequency for electron emission and an absence of lag time, suggested the current understanding of the nature of light was insufficient.
Albert Einstein worked to solve this conundrum. Using Max Planck's theories about how light was carried in "packets", Einstein theorized that light was quantized in discrete particles, which he ended up dubbing as photons. He postulated that the absorption of a quanta of energy is what causes the ejection of an electron. This explained the dependence on frequency instead of intensity that was experimentally observed. Light with a high intensity but only low-energy quanta would not result in an emission. The frequency needed to be high enough- hence the idea of a threshold frequency. Einstein's ideas about the photoelectric effect paved the way for the modern-day interpretation of light's wave-particle duality.
Mechanism and Mathematical Model
When electromagnetic waves are shone onto a surface (often a metal), an electron can be emitted, dependent on the energy of the photons of that light. Electrons emitted in this way are referred to as photoelectrons. The photon transfers energy to the electron, causing it to excite and by ejected from the material. This energy transfer manifests as kinetic energy with the electron. Each photon of light has an energy [math]\displaystyle{ {E=hf} }[/math] where h is Planck's constant and f is the frequency. When the photon contacts the electron on the surface of the material, its energy can be modeled by the relation [math]\displaystyle{ {E_f = K+Φ} }[/math] The quantity Φ is known as the work function of the material and is a unique value for each metal. This can be rewritten as [math]\displaystyle{ {K= hf-Φ} }[/math] to model the kinetic energy. From this, the mathematical reasoning for a threshold frequency can be observed. The minimum kinetic energy the particle can have is 0, it may not be negative. Thus, the frequency must be great enough for K to take on a non-negative value[3].
Significance
The photoelectric effect is significant in that the revelations that stemmed from its observation fundamentally changed the landscape of physics. Einstein's suggestion about the existence of photons and that light has both properties of particles and waves opened the door for an entirely new branch of physics. This realization about the nature of light was then extended to all matter. All matter was theorized then to have wave and particle properties, and thus quantum mechanics was born. Quantum mechanics is now the basis for modern-day physics and has furthered our understanding of astrophysics, electricity, computing, and much more.
Connectedness
The study of the photoelectric effect is one of the catalysts for the formation of quantum mechanics, and quantum mechanics is intimately involved with my interest in cosmology. Understanding the photoelectric effect is a requirement of being a physics major as it is one of the most important phenomena that has been studied in the field.
Applications of the photoelectric effect include photoelectron spectroscopy and night vision technology[4].
Problems
Simple
Question: The photoelectric effect is a common phenomenon, however, it's likely that you haven't noticed it in your everyday life. Explain why that might be.
Answer: Here are two possible explanations. The first is that often when the photoelectric effect occurs, it is difficult to observe due to the low amount of energy that ends up being emitted from the material. The second is that very often when light is shone on a surface, it likely does not have a high enough frequency to reach the threshold frequency for a photoelectron to be emitted. For example, if you shine a flashlight on your laptop-- the photoelectric effect may/may not occur, but either way, you won't be able to observe it with your naked eye.
Middling
Question: An unknown material has a work function value Φ = 2.29 eV and ejects a photoelectron at 8900 m/s. What is the energy of the photon that struck that material?
Answer: The energy relation for the photoelectric effect is given as [math]\displaystyle{ {E_f = K+Φ} }[/math]. The work function value has been given but is in eV. You can give your answer in eV but either way, you will need to do a conversion. The relation between eV and Joules is [math]\displaystyle{ {1 eV = 1.602 * 10^-19 J} }[/math]. You need to calculate the kinetic energy of the photoelectron. At a velocity of 8900 m/s and electron mass of [math]\displaystyle{ {9.11*10^-31}. Using \lt math\gt {K = 0.5mv^2} }[/math], we have that the kinetic energy is [math]\displaystyle{ {3.61 * 10^-23 J} }[/math]. Combining with the work function, we have [math]\displaystyle{ {3.61 * 10^-23 J + (1.602 * 10^-19)(2.29) = E_k} }[/math]. We then have that [math]\displaystyle{ {E_k = 3.66 * 10^-19 J} }[/math].
More Difficult
Question: A beam of electrons is shot at a metal plate with a frequency of 400 Hz and an electron is observed as being emitted with 1.65*10^31 J of energy. Another beam is shot at the plate at 300 Hz. Will the photoelectric effect be observed?
Answer: The problem first requires that the work function for the metal be calculated. The work function is defined as [math]\displaystyle{ {E_f - k = Φ} }[/math]. The energy must be calculated from the frequency using the relation [math]\displaystyle{ {hf = E_f} }[/math]. Using this, you should get [math]\displaystyle{ { E_f = 2.65*10^-31 J} }[/math]. Using this, you should find that [math]\displaystyle{ {Φ = 1 * 10^-31 J} }[/math]. Now applying this to the second beam of electrons and calculating the energy from the given frequency, you should have [math]\displaystyle{ {1.988*10^-31 J - 1 *10^-31 J = K} }[/math]. You can see that the kinetic energy will be positive, and thus may conclude that the photoelectric effect will be observed.
See also
The photoelectric effect is a part of or closely related to all of the articles in the Photons section of the Modern Physics hub of the wiki. These pages include Spontaneous Photon Emission, Quantum Properties of Light, and Electronic Energy Levels and Photons. If interested in the furthering and expansion of these ideas, look to the Quantum Mechanics section.
External links
Photoelectric Effect Phet Simulation
Professor Dave Explains the Photoelectric Effect
References
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