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the_photoelectric_effect [2021/02/01 04:45] – created adminthe_photoelectric_effect [2022/09/01 20:11] (current) – [In Class Activity] admin
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   * Above $\nu_0$, the number of electrons emitted increases with intensity, but does not depend on frequency.   * Above $\nu_0$, the number of electrons emitted increases with intensity, but does not depend on frequency.
   * The kinetic energy of each ejected electron increases linearly with frequency, but does not depend on intensity.   * The kinetic energy of each ejected electron increases linearly with frequency, but does not depend on intensity.
 +
 +{{ :photoelectricgraph.png?direct&400 |}}
 +
 +This is hard to explain with classical physics because:
 +
 +  * The energy density of an electromagnetic wave is proportional to its intensity, so light of any frequency should be able to free an electron if the intensity is large enough.
 +  * If the intensity of the incident radiation is weak, it may take a long time to supply enough energy to free an electron, so why do we see electrons emitted instantly as soon as the light is turned on?
 +
 +===== Einstein's Explanation =====
 +
 +Einstein assumed that light of frequency $\nu$ is //made// of particles, each of which has energy
 +
 +\[\boxed{E = h\nu,}\]
 +
 +which are now called //**photons**//
 +
 +This is more radical than Planck's hypothesis.  Planck assumed that matter and radiation //exchange// energy in discrete packets, but both matter and radiation otherwise obey the laws of classical physics.  Einstein's hypothesis is much more difficult to reconcile with classical physics because how can electromagnetic radiation, which classically is a wave, also be a particle?
 +
 +Einstein assumed that electrons are emitted from the metal due to collisions between the photons and electrons.  Let $W$ be the amount of work required to eject an electron from the metal (its //**work function**//) and $K$ the kinetic energy of an emitted electron.  Then, by conservation of energy
 +
 +\[h\nu = W + K.\]
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 +Obviously, $h\nu$ has to be larger than $W$ in order for an electron to be emitted.  The //**cutoff frequency**// $\nu_0$ is the frequency at which the photon has just enough energy to eject an electron with zero kinetic energy, so $h\nu_0 = W$ or $\nu_0 = W/h$.  Then we have
 +
 +\[K = h(\nu - \nu_0).\]
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 +This matches the observed relationship between $\nu$ and $K$.
 +
 +{{ :photoelectric.png?direct&600 |}}
 +
 +{{:question-mark.png?direct&50|}}
 +====== In Class Activity ======
 +
 +  - We can measure the photoelectric effect by setting up two metal plates with a variable potential difference $V$ between them.  If we shine light of frequency $\nu$ on the positive plate there will be a current between the plates due to the motion of the electron ejected from the positive plate with nonzero kinetic energy, provided $V$ is not too large.  Derive an expression for the //**stopping potential**// $V_s$ at which there ceases to be a current in terms of $h$, $e$, $\nu$, and $W$.
 +
 +{{ :stoppingpotential.png?direct&400 |}}