1.

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

Answer» Here, Z=79,
`r_0=39.5fermi=39.5xx10^(-15)m , E=?`
form `E=1/2mv^2=1/(4pi in_0)((Ze)(2e))/(r_0)`
`E=(9xx10^9xx79xx2(1.6xx10^(-19))^2)/(39.5xx10^(-15))`
`=(18xx79xx2.56)/(39.5xx10^(-14))J`
`=(9.216xx10^(-13))/(1.6xx10^(-13))=5.76MeV`


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