When an alpha particle (a^4 He nucleus) scatters from another nucleus in a very

Question

When an alpha particle (a^4 He nucleus) scatters from another nucleus in a very energetic, head-on collision, the two objects can get quite close together. This so-called "distance of closest approach", r_min, can be determined by assuming that all of the kinetic energy of the incident a is converted into potential energy in the Coulomb field of the target nucleus. What is the distance of closest approach between a 5 MeV alpha and a carbon nucleus (Z = 6, A = 12)? Express all distances in fm (10^-15 m) and all energies in Mev (10^6 eV). As a rough rule-of-thumb, the radius of a nucleus is given by R ~ 1.1 A^1/3 fm, where A is 1/3 the nuclear mass number, the sum of the number of protons and neutrons. If an alpha particle (Z = 2, A = 4) scatters from a gold nucleus (Z = 79, A = 197), what is the minimum kinetic energy an alpha particle needs to experience non-Coulombic nuclear forces, 1.e., to have the two surfaces just "touch" each other? For the lighter nuclei, Z ~ 1/2 A. What nuclear mass number A is required if a 10 MeV alpha is to just reach the nuclear surface?

Explanation / Answer

distance of closest approach=K*q*Q/Ea

q=charge of alpha=2e

Q=charge of gold=6e

Ea=kinetic energy=5*10^6*1.6*10^(-19) J

we got d=3.456*10^(-15) =3.456 fm

(b) d=distance of closest approach=radius of nucleus=1.1*(197)^(1/3)=6.4 fm

so we got Ea=K*q*Q/d=35.54 MeV

(c) for lighter mass

let mass number is A

10*10^6*1.6*10^(-19)=K*2*(A/2)*(1.6*10^(-19))^2/(1.1*(A)^1/3)

we got A=21

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.