an alpha partical accerated by a potential difference v strikes are nucleus .if r○ be the closet distance of a approach of the partical to the nuclus. prove that the (r○=14.4z/v) A
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For α - particle:
2
1
mv
2
=
4πϵ b
q
1
q
2
2eV=
4πϵ
0
r
2e×q
2
q
2
= 4πϵ
0
rV for proton
eV=
4πϵ b
q
1
q
2
eV=
4πϵ b
e q
2
q
2
=4πϵ b V
∴b=r
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