An `alpha`-particle having kinetic energy x MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha`-particle reaches is

Monjur Alahi

An `alpha`-particle having kinetic energy x MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha`-particle reaches is `1.67 xx 10^(-14) m`.
What is the value of x ?
[Given : Atomic number of Cu = 29, K = `9 xx 10^(9) Nm^(2)//C^(2)`]


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Salim Ahmed Kawsar

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Correct Answer - 5
`K.E.=(KZe.2e)/(r ), " "K.E.=(9xx10^(9)xx29xx2(1.6 xx 10^(-19))^(2))/(1.67xx10^(-14))=5Mev`

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