The document discusses an experiment to determine Planck's constant using the photoelectric effect, whereby various wavelengths of light were applied to a metal surface and the emitted photo-electron energies measured. The provided data includes different wavelengths and corresponding stopping potentials, from which calculations lead to a derived value for Planck's constant. The document emphasizes the application of Einstein's photoelectric equation in the analysis of the results.

1. Physics Helpline
L K Satapathy
Modern Physics - 4
-Decay of Nucleus
Photo-electric Effect 2
21
2
hcmv 

 

2. Physics Helpline
L K Satapathy
Modern Physics - 4
Question : In a historical experiment to determine Planck’s constant , a metal
surface was irradiated by different wavelengths. The emitted photo-electron
energies were measured by applying a stopping potential. The relevant data for the
wavelength () of incident light and the corresponding stopping potentials (Vo) are
given below.
34 34 34 34
( ) 6.0 10 ( ) 6.4 10 ( ) 6.6 10 ( ) 6.8 10a b c d   
   
 (m)
V (volt)
0.3 0.4 0.5
2.0 1.0 0.4
Given that and .
Planck’s constant (in units of Js) found from such an experiment is
8
3 10c m s  19
1.6 10e C
 

3. Physics Helpline
L K Satapathy
Modern Physics - 4
Answer :
2
max
1
2 o
hcmv eV

  
2
2
. . . (2)hc eV

 
According to Einstein’s Photo-electric equation
 For the data given in the Q , we can write
1
1
. . . (1)hc eV

 
3
3
. . . (3)hc eV

 

4. Physics Helpline
L K Satapathy
Modern Physics - 4
Correct option = (b)
8 19
7 7
1 13 10 1.6 10 (2 1)
3 10 4 10
h 
 
        
  
 15 19310 1 1.6 10
4
h 
    
 15 19110 1.6 10
4
h 
   
34
6.4 1 [ ]0 nsh J s A
  
1 2
1 2
1 1(1) & (2) ( )hc e V V
 
 
    
 
 15 191 13 10 1.6 10
3 4
h 
     

5. Physics Helpline
L K Satapathy
Modern Physics - 4
Correct option = (b)
2 3
2 3
1 1(2) & (3) ( )hc e V V
 
 
    
 
8 19
7 7
1 13 10 1.6 10 (1 0.4)
4 10 5 10
h 
 
        
  
 15 191 13 10 1.6 10 0.6
4 5
h 
      
34
6.4 1 [ ]0 nsh J s A
  
15 193 310 1.6 10
20 5
h 
     
34 3 201.6 10
5 3
h 
    
Let us test the result using equations (2) & (3)

6. Physics Helpline
L K Satapathy
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