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

Options

A. `26.17 xx 10^6 eV`
B. `26.147738 xx 10^4 eV`
C. `25.3522 xx 10^4 ` eV
D. `20.2254 xx 10^4` eV

Answer

A

Explanation

`r_("nucleus") = 1.3xx10^(-15) xx A^(1//3) m`
`r_u = 1.3 xx 10^(-15) xx (238)^(1//3) = 8.056 xx 10^(-15) m`
`r_(alpha) = 1.3xx10^(-15) xx 4^(1//3) = 2.064 xx 10^(-15) `m
Distance between nucleus of uranium and alpha particle
`r= r_u + r_alpha`
`= 8.056xx10^(-15) + 2.064 xx 10^(-15) m`
`= 10.120 xx 10^(-15)` m
Charge in th nucleus of Uranium `Q_u = 92 xx 1.6 xx 10^(-19) C`
Charge in the nucleus of Alpha `Q_(alpha) = 2 xx 1.6 xx 10^(-19) C`
Repulsion energy `= (Q_u xx Q_(alpha))/(r) xx K_epsi`
`= (92xx1.6xx10^(-19) xx 2 xx 1.6xx 10^(-19))/(10.120xx10^(-15)) xx 9 xx 10^9`
`[K_epsi = 9 xx 10^9] ` coulomb constant
`= 40.54xx10^(15) xx 10^(-19) xx 10^(-19) ` Joule `xx 9 xx 10^(9)`
`= 46.54xx9xx10^(+15) xx 10^(-19) xx 10^(-19) xx 10^9 xx (eV)/(1.6xx10^(-19))`
`= 26.17 xx 10^6` eV