# Identify the missing species and the radioactive decay process?

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#a)# #""_(94)^(239) "Pu" + ""_0^1 "n" -> ""_(Z)^(A) + ""_(36)^(94) "Kr" + 2_(0)^(1) "n"#

#b)# #""_(92)^(233) + ""_(0)^(1) "n" -> ""_(51)^(133) "Sb" + ""_(Z)^(A)? + 3_(0)^(1) "n"#

#c)# #""_(56)^(128) "Ba" + ""_(-1)^(0) "e" -> ""_(Z)^(A) ? + gamma#

#d)# #""_(Z)^(A) ? + ""_(-1)^(0) "e" -> ""_(35)^(81) "Br" + gamma#

##### 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 of#144# , meaning that it is cerium-144,#color(blue)(""_(58)^(144) "Ce")# .

Nuclear fissionis 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

#color(blue)(""_(41)^(98) "Nb")# . This is alsonuclear fissionfor the same reason as in#a# .

The missing product is

#color(blue)(""_(55)^(128) "Cs")# .Since the isotope absorbs an electron to combine with a proton and form a neutron,

#A# stays the same but#Z# decreases by#1# , so I'd call thiselectron capture.

This is just a variation on

#l# . So, it'selectron capture, and the missing reactant is#""_(36)^(81) "Kr"# .