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?

Question

Number of alpha particles N scattered at an angle &#952; during Rutherford&#8217;s alpha scattering experiment is :a)b)c)d)Correct answer is option 'A'. Can you explain this answer?

Answer

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 )

Answer

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 )

Answer

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 )

Answer

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 )

Answer

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 )

Answer

Explain it

Answer

'A'...from book..!

Top Courses for Class 12

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams