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Wave Optics

Question
CBSEENPH12038582

In a Rutherford's α-scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of 60°. What will be the number of scintillations per minute at an angle of 120°?

Solution
Given,
Number of scintillations per minute at an angle 60°, n1 = 8100 m
Number of scintillations per minute at an angle 120° , n2 =? 

The scattering in the Rutherford's experiment is proportional to cot4ϕ2. 

         n2n1 = cot4ϕ2/2cot4ϕ1/2  

Therefore,
 
          n2n1 = cot4120°2cot460°2         = cot 4 60°cot4 30°        = 1334         = 181 
This implies, 

       n2 = 181× n1     = 181×8100     =100.