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..!

Similar Class 12 Doubts

Which of these is true?

Rutherfords experiments on scattering of alpha particles proved that

Which of the following is/are deduced from the Rutherfords scattering experiment?(1) There are neutrons inside the nucleus.(2) The sign of the charge of the nuclei is the same as the sign of alpha particles.(3) Electrons are embedded in the nucleus.

Probability of backward scattering (i.e., scattering of -particles at angles greater than 90) predicted by Thomsons model is

In Rutherford’s experiment, a thin gold foil was bombarded with alpha particles. According to Thomson’s “plum-pudding” model of the atom, what should have happened?

Top Courses for Class 12

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams