Question

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. 

Answer 1

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)
   ^h/_e= gradient
   1.25 - 0.5  
    (8-6) x 10^{14 
   0.75}/₂ 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
   Read from graph = ans
   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