In a scattering experiment, find the distance of closest approach of a 6MeV alpha particle is used.
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Dear Student,
Please find below the solution to the asked query:
The distance of closest approach is given as in under:r0 = 14πε0×2×Z×e2KEgiven KE = 6 MeVand assuming the nucleus to be that of copper. then, Z = 29 and e = 1.6×10−19Csubstituting the values in the above equation and calculating we have,r0 = 2 × 10−14 m
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Please find below the solution to the asked query:
The distance of closest approach is given as in under:r0 = 14πε0×2×Z×e2KEgiven KE = 6 MeVand assuming the nucleus to be that of copper. then, Z = 29 and e = 1.6×10−19Csubstituting the values in the above equation and calculating we have,r0 = 2 × 10−14 m
Hope this information will clear your doubts about (Atom).
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
izerrandom9:
Why do we assume it to be of copper?
Answered by
36
Find the image attached!!
We don't assume it to be copper we know its gold as the question specifies in scattering experiment
For better understanding: potential energy formula used is K(Ze)(2e)/r
Where 2e is the charge of alpha particle and
Z is atomic number of gold
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