Question

An unstable nucleus that is initially at rest decays into a nucleus of fermium-252 containing 100 protons and 152 neutrons and an alpha particle that has a kinetic energy of 8.42 MeV. The atomic masses of helium-4 and fermium-252 are 4.00260 u and 252.08249 u, respectively.

a. What is the atomic number of the original unstable nucleus?

b. What is the velocity of the alpha particle?

c. Where does the kinetic energy of the alpha particle come from? Explain briefly.

d. Assuming all of the energy released in the reaction is in the form of kinetic energy of the alpha particle, determine the exact mass of the original unstable nucleus, to an accuracy of 3 thousandths of a decimal.

e. Suppose that the fermium-252 nucleus could undergo a decay in which a β^– particle was produced. How would this affect the atomic number of the nucleus? Explain briefly.

Answer 1

(a) The reaction can be written as follow: ?  →²⁵²Fm₁₀₀ + ⁴He₂. For nucleons to add up properly, the original nucleus must have been ²⁵⁶X₁₀₂ (this is called Nobelium. FYI)

(b) K = ½ mv² … 8.42 x 10⁶ eV *1.6 x 10^–19 J/eV = ½ (4.0026u*1.66x10^–27 kg) v² … v = 2.014 x 10⁷ m/s

(c) The kinetic energy comes from the conversion of mass to energy in the reaction. The mass before the reaction and the mass after the reaction are unequal, this is known as the mass difference. The energy equivalent of this mass difference contributes to the kinetic energy of the alpha particle.

(d) Converting the alpha particle energy into mass equivalence. 8.42 Mev / (931 MeV /u) = 0.009 u. Adding the masses of all the products. 252.08249 + 4.0026 + 0.009 u = 256.094 u

(e) In B^– decay, a neutron turns into a proton and releases an electron beta particle. Since there is now one more proton, the atomic number increases by 1.