# 1. Theory ### A. Review of Radiation The emission and absorption of light was an early subject for investigation by German physicist Max Planck. As Planck attempted to formulate a theory to explain the spectral distribution of emitted light based on a classical wave model, he ran into considerable difficulty. Classical theory (Rayleigh-Jeans Law) predicted that the amount of light emitted from a black body would increase dramatically as the wavelength decreased, whereas experiment showed that it approached zero. This discrepancy became known as the ultraviolet catastrophe. Experimental data for the radiation of light by a hot, glowing body showed that the maximum intensity of emitted light also departed dramatically from the classically predicted values (Wien's Law). In order to reconcile theory with laboratory results, Planck was forced to develop a new model for light called the quantum model. In this model, light is emitted in small, discrete bundles or quanta. Through a careful study of the photoelectric effect, as we will do in this lab, it is possible to investiage the relationship between the classical and quantum theories of for the emission of light. ### B. The Photo-Electric Effect In photoelectric emission, light strikes a material, causing electrons to be emitted. The classical wave model predicted that as the intensity of incident light was increased, the amplitude and thus the energy of the wave would increase. This would then cause more energetic photoelectrons to be emitted. The new quantum model, however, predicted that higher frequency light would produce higher energy photoelectrons, independent of intensity, while increased intensity would only increase the number of electrons emitted (or photoelectric current). In the early 1900s several investigators found that the kinetic energy of the photoelectrons was dependent on the wavelength, or frequency, and independent of intensity, while the magnitude of the photoelectric current, or number of electrons was dependent on the intensity as predicted by the quantum model. Einstein applied Planck's theory and explained the photoelectric effect in terms of the quantum model using his famous equation for which he received the Nobel prize in 1921: $E = h\nu = KE_{max} + W_{0}$ where $KE_{max}$ is the maximum kinetic energy of the emitted photoelectrons, and $W_{0}$ is the energy needed to remove them from the surface of the material (the work function). E is the energy supplied by the quantum of light known as a photon. ### C. Inverse photoelectric effect (Duane-Hunt experiment) As described in Section 3.7 of Thornton & Rex the reverse process can occur as well. Namely energetic electrons incident on a material will produce radiation (i.e photons), and just like there is a work function in the photo-electric effect (i.e a mininmum photon energy needed to produce photo-electrons) there is a MAXIMUM energy of photons which can be produced, corresponding to a MAXIMUM frequency, $f_{MAX}$. In this experiment we are going to send energetic electrons, through a vacuum, toward a target. When they hit the target, the electrons will slow down….thus they will accelerate (remember “acceleration is a generic term for speeding up or slowing down). This acceleration will result in the production of an EM wave. It is given a special name called “Bremsstrahlung” (i.e. braking radiation, accelerated charged particles give off radiation.) Since the electrons are very energetic, they will produce X-rays. We are trying to test how the max frequency of the x-rays depends on the accelerating voltage. Based on the original Duane-Hunt experiments it was found that $e V_{acc} = h f_{MAX} = \frac{hc}{\lambda_{min}}$ Where $V_{acc}$ is the acceperating voltage of the electrons. ### D. X-ray Spectroscopy X-rays of all sorts of frequencies are formed via bremsstrahlung. But we only want the maximum. To measure this we must separate out the different wavelengths of the x-rays that are created and measure their intensity (because we want to find the minimum wavelength). - How will we separate them? One way to separate them is by using a diffraction grating (just like we did in the photoelectric lab). But because of the very small wavelength of x-rays (order of $10^{-12}$ to $10^{-10}$ m) we would need to have a slit width, in our diffraction grating smaller than the x-rays wavelength! But luckily that is the size of the atom. So, for our diffraction grating, use the atomic crystal structure of a known crystal => NaCl (or something else potentially). Why a crystal? Because we need a regular spacing of our “diffractors” (i.e. atoms) to get a discernable pattern - This means that if our detector is at an angle of $\theta$ , then we will only detect wavelengths that are: $2d\sin{\theta}=\lambda$ where d is the seperation between the atoms. So by positioning the detector at a certain angle, we can get intensity versus wavelength. The computer program used to run our apparatus does this automatically. - Lastly we don’t need to measure the entire spectrum of x-rays. Only find the $f_{MAX}$. So just scan over a small range of angles.