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alpha- particle having 27 MeV has 1.14xx...

`alpha`- particle having 27 MeV has `1.14xx10^(-14)` m distance of closest approach from a nucleus of atom. Calculate atomic no. of atom.

A

100

B

103

C

105

D

90

Text Solution

Verified by Experts

The correct Answer is:
B
PE, at `r_(0)` distance = kinetic energy,
`(k(Ze)(2e))/(r_(0))=27MeV`
`(kxx2Ze^(2))/(r_(0))=27MeV`
`:.Z=(27MeVxxr_(0))/(kxx2e^2)`
`=(27xx10^(6)xx1.1xx10^(-14))/(9xx10^(9)xx2xx1.6xx10^(-19))`
`=1.03125xx10^(2)=103`
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