Question

g The "Coulomb barrier" is defined to be the electric potential energy of a system of two nuclei when their surfaces barely touch. The probability of a nuclear reaction greatly increases if the energy of the system is above this barrier. What is the Coloumb barrier (in MeV) for the absorption of an alpha particle by a lead-208 nucleus

Answer 1

Answer:

The Coulomb Barrier U is 25.91 MeV

Explanation:

Given that:

Atomic Mass of lead nucleus A = 208

atomic mass of an alpha particle A = 4

Radius of an alpha particle $R_\alpha = R_o A^{^{\dfrac{1}{3}}$

where;

$R_\alpha = 1.2 \times 10 ^{-15} \ m$

$R_\alpha = R_o A^{^{\dfrac{1}{3}}$

$R_\alpha = 1.2 \times 10 ^{-15} \ m \times (4) ^{^{\dfrac{1}{3}}$

$R_\alpha = 1.905 \times 10^{-15} \ m$

Radius of the Gold nucleus

$R_{Au}= R_o A^{^{\dfrac{1}{3}}$

$R_{Au}= 1.2 \times 10 ^{-15} \ m \times (208) ^{^{\dfrac{1}{3}}$

$R_{Au} = 7.11 \times 10^{-15} \ m$

$R = R_\alpha + R_{Au}$

$R = 1.905 \times 10^{-15} \ m + 7.11 \times 10^{-15} \ m$

$R = 9.105 \times 10 ^{-15} \ m$

The electric potential energy of the Coulomb barrier $U = \dfrac{Ke \ q_{\alpha} q_{Au}}{R}$

$U = \dfrac{8.99 \times 10^9 \ N.m \ ^2/C ^2 \ \times 2 ( 82) \times \(1.60 \times 10^{-19} C \ \ e } {9.105 \times 10^{-15} \ m }$

U = 25908577.7eV

U = 25.908577 × 10⁶ eV

U =  25.91 MeV

The Coulomb Barrier U is 25.91 MeV