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An alpha particle of energy 5MeV is scattered through 180º by a fixed uranium nucleus. The distance of closest approach is of order of
- a)10-12cm
- b)10-10cm
- c)10-15cm
- d)1Å
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
An alpha particle of energy 5MeV is scattered through 180º by a f...
Distance of closet approx:−
r0=4π∈0Eze2e
KE=PE
21mv2=4π∈01rq1q2
r=5×106×1.6×10−199×109×2×92×(1.6×10−19)2
r=5.3×10−14m≈10−12cm
r0=4π∈0Eze2e
KE=PE
21mv2=4π∈01rq1q2
r=5×106×1.6×10−199×109×2×92×(1.6×10−19)2
r=5.3×10−14m≈10−12cm
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An alpha particle of energy 5MeV is scattered through 180º by a f...
Scattering of an alpha particle involves the interaction of the alpha particle with a target nucleus. When an alpha particle is scattered through 180 degrees, it means that it undergoes a head-on collision with the target nucleus and is reflected back in the opposite direction.
The energy of the alpha particle before scattering is given as 5 MeV. After scattering, the energy of the alpha particle may change due to the interaction with the target nucleus. The change in energy can be calculated using the conservation of energy and momentum.
When an alpha particle is scattered through 180 degrees, the initial and final momenta of the alpha particle are equal in magnitude but opposite in direction. Therefore, the change in momentum of the alpha particle is zero.
Since the change in momentum is zero, the change in kinetic energy of the alpha particle is also zero. Therefore, the energy of the alpha particle remains the same after scattering, which is 5 MeV.
The energy of the alpha particle before scattering is given as 5 MeV. After scattering, the energy of the alpha particle may change due to the interaction with the target nucleus. The change in energy can be calculated using the conservation of energy and momentum.
When an alpha particle is scattered through 180 degrees, the initial and final momenta of the alpha particle are equal in magnitude but opposite in direction. Therefore, the change in momentum of the alpha particle is zero.
Since the change in momentum is zero, the change in kinetic energy of the alpha particle is also zero. Therefore, the energy of the alpha particle remains the same after scattering, which is 5 MeV.
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An alpha particle of energy 5MeV is scattered through 180º by a fixed uranium nucleus. The distance of closest approach is of order ofa)10-12cmb)10-10cmc)10-15cmd)1ÅCorrect answer is option 'A'. Can you explain this answer?
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