Identify the missing species and the radioactive decay process?
a) 23994Pu+10n→AZ+9436Kr+210n
b) 23392+10n→13351Sb+AZ?+310n
c) 12856Ba+0−1e→AZ?+γ
d) AZ?+0−1e→8135Br+γ
1 Answer
For convenience, the original questions are copied down below.
When you set up the system of equations to solve this by asserting conservation of mass, just look across the top and across the bottom to get:
239+1=A+94+2(1)
94+0=Z+36+2(0) Solving each one gives:
A=239+1−94−2=144
Z=94+0−36−2(0)=58 So, your isotope has an atomic number of
58 and a mass number of144 , meaning that it is cerium-144,14458Ce .Nuclear fission is by definition absorbing a neutron and then splitting into two atoms and releasing a couple of neutrons.
Set up the system of equations:
233+1=133+A+3(1)
92+0=51+Z+3(0) Solving these gives:
A=233+1−133−3=98
Z=92+0−51−3(0)=41 Atomic number 41 is niobium, so we have
9841Nb . This is also nuclear fission for the same reason as ina .
The missing product is
12855Cs .Since the isotope absorbs an electron to combine with a proton and form a neutron,
A stays the same butZ decreases by1 , so I'd call this electron capture.
This is just a variation on
l . So, it's electron capture, and the missing reactant is8136Kr .
