In a head-on collision between an alpha-particle and a gold nucleus, the distance of closest approach is 39.5 fermi. Calculate the energy of alpha-particle. For gold Z=79.

In a head-on collision between an alpha-particle and a gold nucleus, t

1.	In a head-on collision between an alpha-particle and a gold nucleus, the distance of closest approach is 39.5 fermi. Calculate the energy of alpha-particle. For gold Z=79.
Answer» Solution :Given `r_(0)=39.5 "fm"=39.5xx10^(-15)m, =79, E=?` As `E=(Ze(2e))/(4 pi epsilon_(0)R)=(2 Ze^(2))/(4 pi epsilon_(0)r)` `=(9xx10^(9)xx79xx2xx(1.6xx10^(-19))^(2))/(39.5xx10^(-15))` `=(9xx79xx2xx1.6xx1.6xx10^(-14))/(39.5)J` or `E=(18xx79xx1.6xx1.6xx10^(-14))/(39.5xx1.6xx10^(-13))` `=5.76` M eV.

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In a head-on collision between an alpha-particle and a gold nucleus, the distance of closest approach is 39.5 fermi. Calculate the energy of alpha-particle. For gold Z=79.

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