Question

1. fill in the table showing

nuclear decay particle	reaction symbol
	$_{-1}^{0}e$
	$_{-1}^{0}e$
positron

1. balance the nuclear equations below.

a. $_{90}^{229}th
ightarrow _{2}^{4}he + _____$
b. $_{8}^{20}o
ightarrow _{9}^{20}f+_____$
c. $_{47}^{104}ag
ightarrow _{-1}^{0}e+_____$
d. $_{51}^{110}sb+_{-1}^{0}e
ightarrow_____$
e. $_{94}^{240}pu
ightarrow _{2}^{4}he+_____$
f. _____$
ightarrow _{-1}^{0}e + _{24}^{54}cr$
g. $_{54}^{118}xe
ightarrow _{+1}^{0}e+_____$
h. $_{80}^{141}nd+_{-1}^{0}e
ightarrow_____$

1. write a balanced nuclear equation for each decay process indicated.

a. the isotope th - 234 decays by an alpha emission.
b. the isotope fe - 59 decays by a beta emission.
c. the isotope c - 11 decays by an electron capture.
d. the isotope tc - 99 decays by a positron emission.
e. the isotope potassium - 40 decays to produce argon - 40.
f. the isotope uranium - 232 decays to produce thorium - 228.
g. the isotope cerium - 144 decays to produce praseodymium - 144.

Explanation:

Step1: Complete nuclear decay particle table

Alpha particle symbol is \( \ce{^4_2He} \); \( \ce{^0_{-1}e} \) is beta-minus; positron symbol is \( \ce{^0_{+1}e} \).

Step2: Balance nuclear equations (a-h)

For each, balance mass/atomic numbers: sum reactants = sum products.

Step3: Write balanced decay equations (a-g)

Identify decay type, apply mass/atomic number conservation.

Answer:

1.

Nuclear Decay Particle	Reaction Symbol
Beta (Beta-minus)	\( \ce{^0_{-1}e} \)
Beta (Beta-minus)	\( \ce{^0_{-1}e} \)
Positron	\( \ce{^0_{+1}e} \)

2.
a. \( \ce{^{225}_{88}Ra} \)
b. \( \ce{^0_{-1}e} \)
c. \( \ce{^{104}_{48}Cd} \)
d. \( \ce{^{110}_{50}Sn} \)
e. \( \ce{^{236}_{92}U} \)
f. \( \ce{^{54}_{23}V} \)
g. \( \ce{^{118}_{53}I} \)
h. \( \ce{^{141}_{79}Au} \)

3.
a. \( \ce{^{234}_{90}Th -> ^4_2He + ^{230}_{88}Ra} \)
b. \( \ce{^{59}_{26}Fe -> ^0_{-1}e + ^{59}_{27}Co} \)
c. \( \ce{^{11}_6C + ^0_{-1}e -> ^{11}_5B} \)
d. \( \ce{^{99}_{43}Tc -> ^0_{+1}e + ^{99}_{42}Mo} \)
e. \( \ce{^{40}_{19}K + ^0_{-1}e -> ^{40}_{18}Ar} \)
f. \( \ce{^{232}_{92}U -> ^4_2He + ^{228}_{90}Th} \)
g. \( \ce{^{144}_{58}Ce -> ^0_{-1}e + ^{144}_{59}Pr} \)