1 \begin{problem*}{28.56}
2 Figure P28.56 shows the stopping potential versus the incident photon
3 frequency for the photoelectric effect for sodupm. Use the graph to
4 find \Part{a} the work function, \Part{b} the ratio $h/e$,
5 and \Part{c} the cutoff wavelength. The data are taken from
6 R.A.~Millikan, \emph{Physical Review} 7:362 (1916).
10 size(5cm, 4cm, IgnoreAspect);
15 pair[] points={(550,0.827),(689,1.20),(741,1.39),
16 (818,1.79),(959,2.29),(1182,3.20)}; // (THz, V)
20 real work_fn = 1.62; // electron-volts
21 real h = 6.56e-34; // joule-seconds
22 real E = h*f*1e12 / 1.6e-19; // electron-volts
27 draw(graph(fit, fmin, fmax), red);
28 draw(graph(points), p=invisible, marker=marker(scale(1pt)*unitcircle, blue));
31 pen thin=linewidth(0.5*linewidth())+grey;
32 xaxis("$f$ (THz)",BottomTop,
33 LeftTicks(begin=false,end=false,extend=true,ptick=thin));
34 yaxis("$\Delta V_s$ (V)",LeftRight,
35 RightTicks(begin=false,end=false,extend=true,ptick=thin));
38 \end{problem*} % problem 28.56
42 The fit line passes nearby the points $(400\U{THz},0\U{V})$ and
43 $(1.20\U{PHz},3.3\U{V})$. In point-slope form, the fit line is then
45 \Delta V_s-0\U{V} &= \frac{(3.3-0)\U{V}}{(1200-400)\U{THz}} (f-400\U{THz}) \\
46 \Delta V_s &= 4.125\text{mV/THz}\cdot (f-400\U{THz})
48 The work function is the inverse $y$-intercept, so
50 \phi = -\Delta V_s(f=0) = -4.125\text{mV/THz}\cdot(-400\U{THz})
55 The theoretical form for the fit line is
57 e\Delta V_s &= hf - \phi \\
58 \Delta V_s &= \frac{h}{e}f - \phi
60 so $\frac{h}{e} = \ans{4.12\U{mV/THz}} = \ans{4.12\U{pV/s}}$.
63 The cutoff wavelength is given by the work function and conservation
66 hf_\text{cut} &= \frac{hc}{\lambda_\text{cut}} = \phi \\
67 \lambda_\text{cut} &= \frac{hc}{\phi} = \ans{751\U{nm}}