Question

A lithium nucleus, while at rest, decays into a helium nucleus of rest mass 6.6483 x 10^–27 kilogram and a proton of rest mass 1.6726 x 10^–27 kilogram, as shown by the following reaction.

In this reaction, momentum and total energy are conserved. After the decay, the proton moves with a speed of 1.95 x 10⁷ m/s.

a. Determine the kinetic energy of the proton.

b. Determine the speed of the helium nucleus.

c. Determine the kinetic energy of the helium nucleus.

d. Determine the mass that is transformed into kinetic energy in this decay.

e. Determine the rest mass of the lithium nucleus.

Answer 1

(a) K = ½ mv² … ½ (1.6726 x 10^–27)(1.95 x 10⁷)² = 3.18 x 10^–13 J

(b) p_before = 0 = p_after → m_pv_p = m_hev_he … (1.6726 x 10^–27)(1.95 x 10⁷) = (6.6483 x 10^–27) v_he … v_he = 4.91 x 10⁶ m/s

(c) K = ½ mv² … ½ (6.6483 x 10^–27)(4.91 x 10⁶)² = 8 x 10^–14 J

(d) The kinetic energy of both of the product particles comes from the conversion of mass in the reaction. The mass equivalent of the total energy of each particle is found with E_total = m c², with E_total the sum of the energies in (a) and (c). The results in a mass equivalent of 4.42 x 10^–30 kg

(e) The mass of the lithium nucleus would be the mass equivalence of the energy above, + the mass of each product resulting in m_Li = 8.3253 x 10^–27 kg