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The Coulomb barrier height for alpha par...

The Coulomb barrier height for `alpha` particle emission is 30.0MeV. What is the barrier height for `._(6)C^(14)`? The required data is
`m(._(88)Ra^(223))=223.01850u , m(._(82)Pa^(209))=208.98107u , m(._(86)Rn^(219))=219.00948u, m(._(6)C^(14))=14.00324u , m(._(2)He^(4))=4.00260u`

Text Solution

Verified by Experts

The coulomb barrier height for a `alpha` decay is equal to the coulomb repulsion between the `alpha` particle and the dougther nucleus of (Z=88-2=86) when they are just touching each other. Hence
`u(alpha)=((2e)(86e))/(r_(0)(219^(1//3)+4^(1//3)))`
Similarly, the coulomb barrier for `C^(14)` is `u(C^(14))= ((6e)(82e))/(r_(0)(209^(1//3)+14^(1//3)))`
Dividing, we get `(u(C^(14)))/(u(alpha))=(6xx82)/(209^(1//3)+14^(1//3))xx(219^(1//3)+4^(1//3))/(2xx86)=(6xx82(6.03+1.59))/((5.93+2.41)xx2xx86)=2.61`
`u(c^(14))=u(alpha)xx2.61=30xx2.61=78.3MeV`
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