Question 24.3: Predicting the Mode of Nuclear Decay Problem Use the atomic ......

Plan If the nuclide is too heavy to be stable (Z > 83), it undergoes α decay. For other cases, we use the Z value to obtain the atomic mass from the periodic table. If the mass number of the nuclide is higher than the atomic mass, the nuclide has too many neutrons: N too high ⇒ β^− decay. If the mass number is lower than the atomic mass, the nuclide has too many protons: Z too high ⇒ β^+ emission and/or e^− capture.

Solution (a) This nuclide has Z = 5, which is boron (B), and the atomic mass is 10.81. The nuclide’s A value of 12 is significantly higher than its atomic mass, so this nuclide is neutron rich. It will probably undergo β^− decay.
(b) This nuclide has Z = 92, so it will undergo α decay and decrease its total mass.
(c) This nuclide has Z = 33, which is arsenic (As), and the atomic mass is 74.92. The A value of 81 is much higher, so this nuclide is neutron rich and will probably undergo β^− decay.
(d) This nuclide has Z = 57, which is lanthanum (La), and the atomic mass is 138.9. The A value of 127 is much lower, so this nuclide is proton rich and will probably undergo β^+ emission and/or e^− capture.

Check To confirm our predictions in parts (a), (c), and (d), let’s compare each nuclide’s N/Z ratio to those in the band of stability: (a) This nuclide has N = 7 and Z = 5, so N/Z = 1.40, which is too high for this region of the band, so it will undergo β^− decay. (c) This nuclide has N = 48 and Z = 33, so N/Z = 1.45, which is too high for this region of the band, so it will undergo β^− decay. (d) This nuclide has N = 70 and Z = 57, so N/Z = 1.23, which is too low for this region of the band, so it will undergo β^+ emission and/or e^− capture. Our predictions based on N/Z values were the same as those based on atomic mass.

Comment Both possible modes of decay are observed for the nuclide in part (d).