PhotoelectricEffect - Maple Help

The Photoelectric Effect

 Overview In this chapter, we explore a transformative experiment that helped pave the way to a new quantum theory: the photoelectric effect. This experiment reveals a new quantum paradigm: the quantization of energy in light.

The Photoelectric Effect

An experiment that was pivotal to the discovery of quantum mechanics was the photoelectric effect, in which electrons can be ejected from the surface of a metal when illuminated with certain frequencies of light.   The electrons at the surface of a metal, being charged and therefore influenced by electric fields, should interact and oscillate along with an incident electric field. According to the classical picture, if the amplitude of the electric field increases, so should the amplitude of electric oscillations until eventually electrons essentially break away from the surface of the metal with a kinetic energy that depends on the intensity of electric field. Furthermore, the photoelectric effect should occur for any frequency, as long as the intensity is sufficiently large.

Unfortunately for classical mechanics, the experimental observations are in complete disagreement with these two predictions. Experimentally, the kinetic energy of the ejected electrons is completely independent of the intensity of incident light, and there exists a threshold frequency, ν${}_{0}$, below which no electrons are ejected, regardless of the intensity! Above the threshold frequency, the kinetic energy of ejected electrons varies linearly with the frequency.  To explain these results, Einstein invoked Planck's energy quantization postulate. Instead of treating only the electric oscillators as quantized, he extended Planck's original idea to suggest that light itself was composed of small, discrete packets of energy, or photons, with E for each photon being proportional to frequency, given by

E = hn.         (4)  Energy of a photon with frequency n

Using Eq. (4)  and the conservation of energy, Einstein showed that

KE = hn - hn0.         (5)  Kinetic Energy of photoelectron

Therefore, a plot of kinetic energy versus frequency should be linear with a slope of h.  Furthermore, the x-intercept is the threshold frequency, ν0.

In this activity, you will determine Planck's constant graphically by fitting simulated photoelectric effect experimental data to Eq. (5). Provided are 4 datasets, Au.dat, Mg.dat, Pb.dat, and Na.dat, each including wavelength (nm) and kinetic energy (eV). Use the Maple input below to determine the value of Planck's constant from each data set. Fill in Table 1 with the threshold frequency, v0, and h from each element's data set and average the values of h.

Table 1: Photoelectric Data for Au, Mg, Pb, and Na

  Au Mg Pb Na Threshold frequency (ν0)     Average value of (h): Planck's constant (h)     

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For each data set, convert units from nm to meters and eV to Joules: (be sure to change the name of the data set in both freq(n) and KE(n)!)

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 ${{\mathrm{_rtable}}}_{{18446744930010061870}}$
 ${{\mathrm{_rtable}}}_{{18446744930010061750}}$ (2.1)

Plot KE versus frequency:

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Finally, fit the data to Eq. (5) using the interactive CurveFitting command. Choose 'Least Squares' then 'Plot' then 'Done'.

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 ${f}{≔}\left(\right)$ (2.2)
 > $B≔\mathrm{plot}\left(f,\mathrm{nu}=1.0ⅇ15..4.1ⅇ15\right):\mathrm{plots}\left[\mathrm{display}\right]\left(A,B\right);$

The graphical determination of Planck's constant, h, is the slope of the line. Enter this into Table 1. The threshold frequency is ν${}_{0}$ = − (y-intercept) / h. Calculate and enter this value into Table 1.

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 ${\mathrm{nu_0}}{≔}{-}\frac{{\mathrm{input}}{}{y}{-}{\mathrm{intercept}}}{{\mathrm{input}}{}{\mathrm{slope}}}$ (2.3)

Now repeat the above calculations for each element in Table 1.

What is the average value for h? How does this compare to the iterative value?

For which element is it most difficult to eject a photoelectron? Why?

 Answer According to the data provided, Au has a larger threshold frequency. Therefore, it is more difficult to remove an electron from Au since a higher energy photon is required.