Identify the missing species and the radioactive decay process?

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Identify the missing species and the radioactive decay process?

$a)$ $a)$ $_{94}^{239} Pu +_{0}^{1} n \to_{Z}^{A} +_{36}^{94} Kr + 2_{0}^{1} n$ $""_(94)^(239) "Pu" + ""_0^1 "n" -> ""_(Z)^(A) + ""_(36)^(94) "Kr" + 2_(0)^(1) "n"$

$b)$ $b)$ $_{92}^{233} +_{0}^{1} n \to_{51}^{133} Sb +_{Z}^{A} ? + 3_{0}^{1} n$ $""_(92)^(233) + ""_(0)^(1) "n" -> ""_(51)^(133) "Sb" + ""_(Z)^(A)? + 3_(0)^(1) "n"$

$c)$ $c)$ $_{56}^{128} Ba +_{- 1}^{0} e \to_{Z}^{A} ? + γ$ $""_(56)^(128) "Ba" + ""_(-1)^(0) "e" -> ""_(Z)^(A) ? + gamma$

$d)$ $d)$ $_{Z}^{A} ? +_{- 1}^{0} e \to_{35}^{81} Br + γ$ $""_(Z)^(A) ? + ""_(-1)^(0) "e" -> ""_(35)^(81) "Br" + gamma$

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Answer 1 · 2017-01-07T22:21:57

For convenience, the original questions are copied down below.

$a)$ $a)$ $_{94}^{239} Pu +_{0}^{1} n \to_{Z}^{A} +_{36}^{94} Kr + 2_{0}^{1} n$ $""_(94)^(239) "Pu" + ""_0^1 "n" -> ""_(Z)^(A) + ""_(36)^(94) "Kr" + 2_(0)^(1) "n"$

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:

$239 + 1 = A + 94 + 2 (1)$ $239 + 1 = A + 94 + 2(1)$
$94 + 0 = Z + 36 + 2 (0)$ $94 + 0 = Z + 36 + 2(0)$

Solving each one gives:

$A = 239 + 1 - 94 - 2 = 144$ $A = 239 + 1 - 94 - 2 = 144$
$Z = 94 + 0 - 36 - 2 (0) = 58$ $Z = 94 + 0 - 36 - 2(0) = 58$

So, your isotope has an atomic number of $58$ $58$ and a mass number of $144$ $144$ , meaning that it is cerium-144, $_{58}^{144} Ce$ $color(blue)(""_(58)^(144) "Ce")$ .

Nuclear fission is by definition absorbing a neutron and then splitting into two atoms and releasing a couple of neutrons.

$b)$ $b)$ $_{92}^{233} +_{0}^{1} n \to_{51}^{133} Sb +_{Z}^{A} ? + 3_{0}^{1} n$ $""_(92)^(233) + ""_(0)^(1) "n" -> ""_(51)^(133) "Sb" + ""_(Z)^(A)? + 3_(0)^(1) "n"$

Set up the system of equations:

$233 + 1 = 133 + A + 3 (1)$ $233 + 1 = 133 + A + 3(1)$
$92 + 0 = 51 + Z + 3 (0)$ $92 + 0 = 51 + Z + 3(0)$

Solving these gives:

$A = 233 + 1 - 133 - 3 = 98$ $A = 233 + 1 - 133 - 3 = 98$
$Z = 92 + 0 - 51 - 3 (0) = 41$ $Z = 92 + 0 - 51 - 3(0) = 41$

Atomic number 41 is niobium, so we have $_{41}^{98} Nb$ $color(blue)(""_(41)^(98) "Nb")$ . This is also nuclear fission for the same reason as in $a$ $a$ .

$c)$ $c)$ $_{56}^{128} Ba +_{- 1}^{0} e \to_{Z}^{A} ? + γ$ $""_(56)^(128) "Ba" + ""_(-1)^(0) "e" -> ""_(Z)^(A) ? + gamma$

The missing product is $_{55}^{128} Cs$ $color(blue)(""_(55)^(128) "Cs")$ .

Since the isotope absorbs an electron to combine with a proton and form a neutron, $A$ $A$ stays the same but $Z$ $Z$ decreases by $1$ $1$ , so I'd call this electron capture.

$d)$ $d)$ $_{Z}^{A} ? +_{- 1}^{0} e \to_{35}^{81} Br + γ$ $""_(Z)^(A) ? + ""_(-1)^(0) "e" -> ""_(35)^(81) "Br" + gamma$

This is just a variation on $l$ $l$ . So, it's electron capture, and the missing reactant is $_{36}^{81} Kr$ $""_(36)^(81) "Kr"$ .

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Identify the missing species and the radioactive decay process?

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