1.

Consider an alpha- particle just in contact with a ._(92)U^(238) nucleus. Calculate the coulombic repulsion energy (i.e., the height of the coulombic barrier between U^(238) and alpha particle) assuming that the distance between them is equal to the sum of their radii

Answer»

`23.851 xx 10^(4) eV`
`26.147738 xx 10^(4) eV`
`25.3522 xx 10^(4) eV`
`20.2254 xx 10^(4) eV`

Solution :`r_("nucleus") = 1.3 xx 10^(-13) xx (A)^(1//3)`, where A is mass number
`r_(U^(238)) = 1.3 xx 10^(-13) xx (238)^(1//8) = 8.06 xx 10^(-13) cm`
`r_(He^(4)) = 1.3 xx 10^(-13) xx (4)^(1//3) = 2.06 xx 10^(-13) cm`
`:.` TOTAL DISTANCE in between URANIUM and `alpha` nuclei
`= 8.06 xx 10^(-13) + 2.06 xx 10^(-13) = 10.12 xx 10^(-13) cm`
Now repulsion energy
`= (Q_(1)Q_(2))/(r) = (92 xx 4.8 xx 10^(-10) xx 2 xx 4.8 xx 10^(-10))/(10.12 xx 10^(-13)) erg`
`= 418.9 xx 10^(-7) erg = 418.9 xx 10^(-7) xx 6.242 xx 10^(11) eV`
`= 26.147738 xx 10^(4) eV`


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