Problem#1
A 4.78-MeV alpha particle from a 226Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?
Answer:
The kinetic energy of the alpha particle is all converted to electrical potential energy at closest
approach. The force on the alpha particle is the electrical repulsion of the nucleus.
The electrical potential energy of the system is
U = kq1q2/r
and the kinetic energy is K = ½ mv2. The electrical force is R = 2.5 m (at closest approach)
(a) Equating the initial kinetic energy and the final potential energy and solving for the separation radius r gives
r = kq1q2/K = (9 x 109 Nm2/C2)(92e)(2e)/[4.78 x 106 Ev)(1.60 x 10-19 J/eV)] = 5.54 x 10-14 m
(b) The above result may be substituted into Coulomb’s law. Alternatively, the relation between the
magnitude of the force and the magnitude of the potential energy in a Coulomb field is
F = |U|/r and |U| = K, so
F = K/r = (4.78 x 106 eV)(1.6 x 10-19 J/eV)/(5.54 x 10-14 m) = 13.8 N
Problem#2
A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in “head-on” to a particular lead nucleus and stops 6.50 x 10-14 m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 x 10-27 kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?
Answer:
(a) If the particles are treated as point charges, U = kq1q2/r
q1 = 2e (alpha particle); q2 = 82e (gold nucleus); r is given so we can solve for U.
U = (8.987 x 109 N.m2/C2)(2)(82)(1.602 x 10-19 C)2/(6.50 x 10-14 m) = 5.82 x 10-13 J
U = (5.82 x 10-13 J)(1 eV/1.602 x 10-19 J) = 3.63 x 106 eV = 3.63 MeV
(b) Apply conservation of energy: K1 + U1 = K2 + U2.
Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle
momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies r1 ≈ ∞ and U1 = 0. Alpha particle stops implies K2 = 0.
Conservation of energy thus says
K1 = U1 = 5.82 x 10-13 J = 3.63 MeV
(c) K = ½ mv2
5.82 x 10-13 J = ½ (9.11 x 10-31 kg)v2
v = 1.32 x 107 m/s
v/c = 0.044, so it is ok to use the nonrelativistic expression to relate K and v. When the alpha
particle stops, all its initial kinetic energy has been converted to electrostatic potential energy.
Post a Comment for "The Nuclear Atom and Atomic Spectra Problems and Solutions"