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प्रश्न
Determine the distance of the closest approach when an alpha particle of kinetic energy 3.95 MeV approaches a nucleus of Z = 79, stops and reverses its directions.
उत्तर
Let 'r' be centre to centre distance between α-particle when α particle is at the stopping point then K = `1/(4piepsilon_0) ((Ze)(2e))/r`
Given: K = 3.95 MeV
= `3.95 xx 10^6 xx 1.6 xx 10^-19` J
⇒ `r = 1/(4piepsilon_0) (2Ze^2)/K`
= `(9 xx 10^9 xx 2 xx 79 xx (1.6 xx 10^-19)^2)/(3.95 xx 10^6 xx 1.6 xx 10^-19)`
= `(9 xx 2 xx 79 xx 1.6 xx 10^-19 xx 10^9)/(3.95 xx 10^6)`
= `576 xx 10^-16` m
`r = 5.76 xx 10^-14` m
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