# Figure 6 shows a graph of stopping potential against the frequency for a certain photo emissive surface, drawn by a student from the data collected when carrying out an experiment on photoelectric effect.

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Figure 6 shows a graph of stopping potential against the frequency for a certain photo emissive surface, drawn by a student from the data collected when carrying out an experiment on photoelectric effect.

From the graph, determine the:

1. threshold frequency of the surface:
2. plank's constant h, given that the energy of the incident photon is
1.6X 10-19 J;
3. work function of the surface.

1. eVs = hf - hfo
at Vs = 0
hf = hfo
f = fo which is obtained by extrapolating the graph to obtain the √ value of fo when Vs=0
4.6 x 1014 1 +2 ± 0.1

2. Vs =  hf/e  -  hfo/e
h/e (f-fo)
1.25 - 0.5
(8-6) x 1014
0.75/x 10-14
0.375 x 10-14
h= 3.75 x 10-15 x 1.6 x 10-19
6.0 x 10-34 Js ± (0.8)
alternatively,
E= hf
h=E/f
1.6 x 10-19
f
f = ( 4.6 x 1014 - 10 x 1014)
Range ( 1.6 x 10-34 - 3.478 x 10-34) Js
3. Wo = hfo
= 6.0x10-34x4.36x1014
=2.76x10-19J