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divider"></i></li><li><a href="../1132/ahl-quantum.html">AHL Quantum</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Photoelectric effect and photons</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 25 minutes"><i class="fa fa-clock-o"></i> 25'</span> </ol> <article id="main-article"> <p><img alt="" src="../../quantum/pv-cells.jpg" style="float: left; width: 250px; height: 166px;">The photoelectric effect is indisputable evidence for the particle nature of light. Particles of light are discrete and named photons. The energy of a photon depends not on intensity but frequency.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Photoelectric effect</p> </div> </div> <div class="panel-body"> <div> <p>There is a significant body of evidence for the wave model of light including interference and diffraction. Reflection and refraction are also wave properties, but are not exclusive to waves.</p> <p>The photoelectric effect is evidence only for the quantum nature of light.</p> <p>When light is shone on a metal plate, electrons in the plate are excited and may escape the surface (photoelectrons). The results of the experiment are:</p> <ul> <li>Below a certain frequency of light, <strong>no</strong> electrons are emitted, regardless of the intensity of the light. This is evidence against the wave theory (in which continuous waves would eventually build up sufficient energy for escape).</li> <li>When the frequency is increased (i.e. towards or into the ultraviolet region of the electromagnetic spectrum), a <b>threshold frequency<em> </em></b>is reached at which electrons are emitted. These are the electrons closest to the surface of the metal. The energy required to just release the electrons is the <strong>work function</strong>. This observation indicates that a single particle of light gives its entire energy to a single electron.</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/376909884"></iframe></div> <ul> <li>Above the threshold frequency, increasing the frequency gives the electrons a higher <strong>maximum</strong> kinetic energy. The range of kinetic energies comes from the variation in where the electrons were located in the metal. This graph shows how the energy of the photoelectrons varies with frequency; none are emitted before the threshold frequency but then kinetic energy increases linearly with frequency.</li> </ul> <p style="text-align: center;"><img alt="" src="../../quantum/photoelectric_effect_diagram.svg" style="width: 400px; height: 269px;"></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/376909914"></iframe></div> <ul> <li>Above the threshold frequency, increasing the intensity increases the <strong>number</strong> of electrons released in a given time, but not the maximum kinetic energy.</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/376909857"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Photons</p> </div> </div> <div class="panel-body"> <div> <p>Photons are discrete packets of electromagnetic energy. The higher the frequency of the electromagnetic energy, the higher the energy of the photon:</p> <p style="text-align: center;"><span class="math-tex">\(E=hf\)</span></p> <ul> <li><span class="math-tex">\(E\)</span> is photon energy (J)</li> <li><span class="math-tex">\(h\)</span> is Planck's constant (6.63×10<sup>−</sup><sup>34</sup> Js)</li> <li><span class="math-tex">\(f\)</span> is frequency (Hz)</li> </ul> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Einstein's equation</p> </div> </div> <div class="panel-body"> <div> <p>Einstein considered the conservation of energy during the photoelectric effect and got a Nobel prize as a result. The energy of the incoming photon becomes the energy required to release the electron and the electron's kinetic energy. When the electron requires only the work function to be released, the kinetic energy is maximum:</p> <p style="text-align: center;"><span class="math-tex">\(hf=hf_0+E_\text{max}\)</span></p> <p style="text-align: center;"><span class="math-tex">\(hf=\Phi+E_\text{max}\)</span></p> <ul> <li><span class="math-tex">\(f_0\)</span> is threshold frequency for a given material (Hz)</li> <li><span class="math-tex">\(\Phi\)</span> is work function for a given material (J)</li> <li><span class="math-tex">\(E_\text{max}\)</span> is maximum kinetic energy</li> </ul> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Wave particle duality</p> </div> </div> <div class="panel-body"> <div> <p>The main deficiency in the wave model of light for explaining the photoelectric effect is that an increased intensity should increase the energy of the electrons. Over time, enough energy would be built up for electrons to be released. And the intensity should increase the maximum kinetic energy of the electrons.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/376909984"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Millikan's experiment</p> </div> </div> <div class="panel-body"> <div> <p>Millikan's experiment enabled the determination of Planck's constant. Millikan placed the irradiated metal plate in a circuit with a vacuum gap. The vacuum gap meant that only photoelectrons could provide a current. For a given frequency of light, a potential difference could be supplied to repel the photoelectrons such that those with the maximum kinetic energy were <i>just </i>prevented from reaching across the gap.</p> <p>The kinetic energy lost by an electron was equal to the potential energy gained in the uniform field:</p> <p style="text-align: center;"><span class="math-tex">\(E_\text{max}=eV\)</span></p> <ul> <li><span class="math-tex">\(e\)</span> is the charge on an electron (C)</li> <li><span class="math-tex">\(V\)</span> is the stopping potential (V)</li> </ul> <p style="text-align: center;"><span class="math-tex">\(hf=\Phi+eV\)</span></p> <p>A graph could, for example, be plotted of stopping potential against frequency of light to obtain Planck's constant.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/376909841"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Energetic interactions with matter</p> </div> </div> <div class="panel-body"> <div> <p>The photoelectric effect is a relatively low energy interaction between photons and matter. What about medium and high energy interactions?</p> <h4>Compton scattering</h4> <p>When the energy of a photon far exceeds the energy required to release an electron (e.g. X-ray frequencies), the electron can be approximated as being free. When the photon collides with the electron, the electron gains energy and momentum; both of these quantities are conserved in the collision.</p> <p>This indicates that the photon had momentum, a property normally associated with matter because of its mass. The photon's loss of energy can be observed in its lower frequency (and increased wavelength).</p> <h4>Pair production and annihilation</h4> <p>The photons of highest energy (i.e. gamma frequencies) have sufficient energy that it can be converted into the mass of particles, according to <span class="math-tex">\(E=mc^2\)</span>. Alternatively, particle mass can be converted into electromagnetic radiation.</p> <p>Pair production occurs when a photon passes near to a nucleus. This condition is necessary to conserve both mass-energy and momentum. The energy of the photon must match or exceed the combined mass of a particle and its antiparticle, for example an electron and positron. This also satisfies conservation of lepton number and charge.</p> <p>The opposite, when an electron and positron meet and annihilate, produces two gamma photons. This process also conserves charge, lepton number, mass-energy and momentum, the latter because of the opposite directions of the photons.</p> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use quizzes to practise application of theory.</em></p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#047260e6"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="047260e6"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> The photoelectric effect <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-358-1135" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.</p><p style="text-align: center;"><img alt="" height="258" src="../../photoe3.png" width="220"></p><p>Planck's constant in Js is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Ae\over C\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Ae\over D\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Be\over D\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(Be\over C\)</span></span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(h\)</span> is the gradient. The only of these options to do is <span class="math-tex">\(B\over C\)</span>.</p><p>The units of the energy axis are eV, so we must multiply by <span class="math-tex">\(e\)</span> to obtain Planck's constant in the required units.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.</p><p style="text-align: center;"><img alt="" height="258" src="../../photoe3.png" width="220"></p><p>The work function in eV is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(C\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(B\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Be\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Ce\)</span></span></label> </p></div><div class="q-explanation"><p>The work function, the minimum energy that must be overcome, is the vertical intercept. The units are already eV.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.</p><p style="text-align: center;"><img alt="" height="258" src="../../photoe3.png" width="220"></p><p>The maximum wavelength for electron emission is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(D\over c\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Cc\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(c\over C\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(c\over D\)</span></span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(C\)</span> is the minimum frequency needed. The equivalent wavelength, <span class="math-tex">\({c\over f}={c\over C}\)</span>.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph represents the maximum kinetic energy of photoelectrons against the frequency of illuminating light.</p><p style="text-align: center;"><img alt="" height="258" src="../../photoe3.png" width="220"></p><p>The work function in eV is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(CD\over B-A\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(AC\over D\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(AC\over D-C\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(AD\over C\)</span></span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(\Phi =hf_0 = \text{gradient} \times C\)</span></p><p>Other methods are possible, but this is the only correct response of these options.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two sources of light, one red and one blue, have the same intensity.</p><p>The ratio <span class="math-tex">\(\text{number of photons from the blue source}\over \text{number of photons from the red source}\)</span> is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>not possible to determine</span></label> </p><p><label class="radio"> <input type="radio"> <span><1</span></label> </p><p><label class="radio"> <input type="radio"> <span>>1</span></label> </p></div><div class="q-explanation"><p>It is the number of photons <u>per unit area</u> that is greater for the red light. We are not told the surface area of the light sources.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Photoelectric emission takes place when a green light source is shone on a sodium plate but not for yellow light. If the green light is replaced by a white light of the same intensity there will be:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>fewer electrons</span></label> </p><p><label class="radio"> <input type="radio"> <span>more electrons</span></label> </p><p><label class="radio"> <input type="radio"> <span>no electrons</span></label> </p><p><label class="radio"> <input type="radio"> <span>the same number of electrons</span></label> </p></div><div class="q-explanation"><p>White light is made of all wavelengths. Only those corresponding to green light or shorter will emit electrons.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Photoelectric emission takes place when a green light source is shone on a sodium plate but not for yellow light. If the green light is replaced by a blue light of the same intensity there will be:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>fewer electrons</span></label> </p><p><label class="radio"> <input type="radio"> <span>no electrons</span></label> </p><p><label class="radio"> <input type="radio"> <span>the same number of electrons</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>more electrons</span></label> </p></div><div class="q-explanation"><p>Although the blue light has fewer photons per unit area they have more energy. More electrons will gain enough energy for emission.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Photoelectric emission takes place when a green light source is shone on a sodium plate but not for yellow light. If the wavelength of the green source is reduced while keeping the intensity constant, the number of electrons emitted will:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>stay the same</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>increase to a maximum then decrease</span></label> </p><p><label class="radio"> <input type="radio"> <span>decrease</span></label> </p><p><label class="radio"> <input type="radio"> <span>increase</span></label> </p></div><div class="q-explanation"><p>Electron current will reach saturation then reduce as maintaining the intensity beyond saturation will reduce the number of photons.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A photocell has a threshold frequency in the green region of the spectrum. To get current to flow for red light could be achieved with:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>higher intensity</span></label> </p><p><label class="radio"> <input type="radio"> <span>higher potential difference</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>none of the other options</span></label> </p><p><label class="radio"> <input type="radio"> <span>more photons</span></label> </p></div><div class="q-explanation"><p>The frequency of red light is below the threshold frequency and so no electrons would have sufficient energy to be emitted</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Light is shone on a metal plate at the threshold frequency for photoelectric emission.</p><p>The ratio of <span class="math-tex">\(\text{photoelectrons} \over \text{photons}\)</span> is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0</span></label> </p><p><label class="radio"> <input type="radio"> <span>>1</span></label> </p><p><label class="radio"> <input type="radio"> <span>1</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><1</span></label> </p></div><div class="q-explanation"><p>Not all of the photons cause photoelectric emission as some electrons will be far beneath the surface</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button 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