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Number of alpha particles N scattered at an angle θ during Rutherford’s alpha scattering experiment is :
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Number of alpha particles N scattered at an angle θ during Ruther...
Answer :- a
Solution :- For a single scatterer, such as a single gold nucleus within a thin gold foil layer, the differential scattering cross section is defined as follows [2]:
dσ(θ, φ)/dΩ = flux scattered into element dΩ at angles (θ, φ)/incident flux per unit area
dσ/dΩ = (Iθ × A)/ (dΩ × I0 × NAvo × ρ × x(foil))........... (1)
where NAvo is Avogadro’s number, xfoil is the thickness of the target foil, A is the atomic mass of the material in the target foil, dΩ is the solid-angle of the detector, I0 is the unattenuated intensity of the alpha particle beam.
dσ/dΩ = [(ZZ0 e^2/4E )^2]/(1/sin^4(θ/2 ))
dσ/dΩ is directly proportional to 1/sin^4(θ/2 )
Most Upvoted Answer
Number of alpha particles N scattered at an angle θ during Ruther...
Answer :- a
Solution :- For a single scatterer, such as a single gold nucleus within a thin gold foil layer, the differential scattering cross section is defined as follows [2]:
dσ(θ, φ)/dΩ = flux scattered into element dΩ at angles (θ, φ)/incident flux per unit area
dσ/dΩ = (Iθ × A)/ (dΩ × I0 × NAvo × ρ × x(foil))........... (1)
where NAvo is Avogadro’s number, xfoil is the thickness of the target foil, A is the atomic mass of the material in the target foil, dΩ is the solid-angle of the detector, I0 is the unattenuated intensity of the alpha particle beam.
dσ/dΩ = [(ZZ0 e^2/4E )^2]/(1/sin^4(θ/2 ))
dσ/dΩ is directly proportional to 1/sin^4(θ/2 )
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Number of alpha particles N scattered at an angle θ during Ruther...
Answer :- a
Solution :- For a single scatterer, such as a single gold nucleus within a thin gold foil layer, the differential scattering cross section is defined as follows [2]:
dσ(θ, φ)/dΩ = flux scattered into element dΩ at angles (θ, φ)/incident flux per unit area
dσ/dΩ = (Iθ × A)/ (dΩ × I0 × NAvo × ρ × x(foil))........... (1)
where NAvo is Avogadro’s number, xfoil is the thickness of the target foil, A is the atomic mass of the material in the target foil, dΩ is the solid-angle of the detector, I0 is the unattenuated intensity of the alpha particle beam.
dσ/dΩ = [(ZZ0 e^2/4E )^2]/(1/sin^4(θ/2 ))
dσ/dΩ is directly proportional to 1/sin^4(θ/2 )
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Number of alpha particles N scattered at an angle θ during Rutherford’s alpha scattering experiment is :a)b)c)d)Correct answer is option 'A'. Can you explain this answer? for Class 12 2026 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Number of alpha particles N scattered at an angle θ during Rutherford’s alpha scattering experiment is :a)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Number of alpha particles N scattered at an angle θ during Rutherford’s alpha scattering experiment is :a)b)c)d)Correct answer is option 'A'. Can you explain this answer?.
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