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

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

where;

Radius of the Gold nucleus

The electric potential energy of the Coulomb barrier

U = 25908577.7eV

U = 25.908577 × 10⁶ eV

U =  25.91 MeV

The Coulomb Barrier U is 25.91 MeV

20 drops hope this

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :