In a Geiger-Marsden experiment, calculate energy of `alpha` particle whose distance of closet approach to the nucleus of Z=79 is `2.8 xx 10^(-14)` m. How will the distance of closet approach be affected when the kinetic energy of the `alpha`-particle is doubled ?
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Here, `r_0=2.8xx10^(-14)m ,Z=79` E=? form `E=1/(4pi in_0)=((Ze)(2e))/(r_0)` `E=1/(4pi in_0)(2Ze^(2))/(r_0)=(9xx10^9xx2xx79(1.6xx10^(-19))^2)/(2.8xx10^(-14))` `1.300xx10^(-12)J` `=(1.300xx10^(-12))/(1.6xx10^(-13))MeV=8.125MeV` As E is doubled, `r_0` becomes half `(r_0)/2=1.4xx10^(-14)m`
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