Question

iii. practice with equations
try these.

1. $_{27}^{60}co\

ightarrow\boxed{\text{(purple - written)} _{28}^{60}ni}+_{ - 1}^{0}e$

1. $_{43}^{99}tc\

ightarrow\boxed{\text{(empty)}}+_{1}^{0}\beta$

1. $_{53}^{131}i\

ightarrow\boxed{\text{(empty)}}+_{ - 1}^{0}e$

1. $\boxed{\text{(empty)}}\

ightarrow_{56}^{137}ba+_{ - 1}^{0}e$

1. $\boxed{\text{(empty)}}\

ightarrow_{11}^{23}na+_{1}^{0}\beta$

1. $_{90}^{229}th\

ightarrow\boxed{\text{(empty)}}+_{2}^{4}he$

Explanation:

Step1: Balance mass - number

In a nuclear - decay equation, the sum of mass - numbers on the left side must equal the sum of mass - numbers on the right side. And the sum of atomic numbers on the left side must equal the sum of atomic numbers on the right side.

Step2: Solve for unknown in 1

For $_{27}^{60}Co
ightarrow X + _{ - 1}^{0}e$, let the unknown be $_{Z}^{A}X$. Mass - number: $A = 60$, atomic number: $Z-1 = 27$, so $Z = 28$. The element with $Z = 28$ is nickel (Ni), so the unknown is $_{28}^{60}Ni$.

Step3: Solve for unknown in 2

For $_{43}^{99}Tc
ightarrow X+_{1}^{0}\beta$, let the unknown be $_{Z}^{A}X$. Mass - number: $A = 99$, atomic number: $Z + 1=43$, so $Z = 42$. The element with $Z = 42$ is molybdenum (Mo), so the unknown is $_{42}^{99}Mo$.

Step4: Solve for unknown in 3

For $_{53}^{131}I
ightarrow X + _{ - 1}^{0}e$, let the unknown be $_{Z}^{A}X$. Mass - number: $A = 131$, atomic number: $Z-1 = 53$, so $Z = 54$. The element with $Z = 54$ is xenon (Xe), so the unknown is $_{54}^{131}Xe$.

Step5: Solve for unknown in 4

For $X
ightarrow_{56}^{137}Ba + _{ - 1}^{0}e$, let the unknown be $_{Z}^{A}X$. Mass - number: $A = 137$, atomic number: $Z-1 = 56$, so $Z = 57$. The element with $Z = 57$ is lanthanum (La), so the unknown is $_{57}^{137}La$.

Step6: Solve for unknown in 5

For $X
ightarrow_{11}^{23}Na+_{1}^{0}\beta$, let the unknown be $_{Z}^{A}X$. Mass - number: $A = 23$, atomic number: $Z + 1=11$, so $Z = 10$. The element with $Z = 10$ is neon (Ne), so the unknown is $_{10}^{23}Ne$.

Step7: Solve for unknown in 6

For $_{90}^{229}Th
ightarrow X+_{2}^{4}He$, let the unknown be $_{Z}^{A}X$. Mass - number: $A = 229 - 4=225$, atomic number: $Z=90 - 2 = 88$. The element with $Z = 88$ is radium (Ra), so the unknown is $_{88}^{225}Ra$.

Answer:

1. $_{28}^{60}Ni$
2. $_{42}^{99}Mo$
3. $_{54}^{131}Xe$
4. $_{57}^{137}La$
5. $_{10}^{23}Ne$
6. $_{88}^{225}Ra$