Question

A) Answer as an isotope

1. Write the identity of the missing nucleus for the following nuclear decay reaction:

232 90Th→228 88Ra+?

2. Write the identity of the missing nucleus for the following nuclear decay reaction:

224 88Ra→42He+?

3. Write the identity of the missing nucleus for the following nuclear decay reaction:

?→237 93Np+42He+00γ

B) Write a balanced nuclear equation for the beta decay of each:

1. 25/11Na

2. 20/8O

3. strontium-92

4.iron-60

C)

1. Complete the following nuclear equation

116C→115B+−−−

2. Complete the following nuclear equation

3516S→−−−+0−1e

3. Complete the following nuclear equation

−−−→9039Y+0−1e

4. Complete the following nuclear equation

21083Bi→ −−−+42He

5. Complete the following nuclear equation

−−−→ 8939Y+0+1e

Answer 1

1)

We have to equate the atomic number and mass number

232 = 228 + x

x = 232-228

x = 4

Atomic number = 90 = 88 + x

x = 90-88

x =2

So missing nucleus is He which has atomic number = 2 and mass number = 4

2) ²²⁴Ra₈₈→⁴He₂ + ?

Missing mass number = x = 224-4 = 220

Mssing atomic number = x = 88-2 = 86

identify the atom using periodic table

²²⁴Ra₈₈→⁴He₂ + ²²⁰Rn₈₆

3) ? → ²³⁷Np₉₃ + ⁴He₂ + ⁰₀γ

Here reactant is missing so add massnumbers of all products and atomic number of all products

Atomic number = 93 +2 =95

Mass number = 237 + 4 = 241

So the atom is = Am

²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂ + ⁰₀γ

B) beta decay is called electron loss which increase the atomic number of the product

1) ²⁵Na₁₁ --------> ²⁵Mg₁₂ + _-1e⁰

Na hass atomic number 11

atomic number increases to 12 with beta decay

2) ²⁰O₈ --------> ²⁰F₉+  _-1e⁰

3) strontium-92

Atomic number of Sr is = 38

⁹²Sr₃₈ --------> ⁹²F₃₉ +  _-1e⁰

4)iron-60

Atomic number of Fe = 26

⁶⁰Fe₂₆ -------->⁶⁰F₂₇ +  _-1e⁰

C) ¹¹₆C → ¹¹₅B + ?

6 = 5 + x

x = 6-5 = 1

No change in mass number

¹¹₆C → ¹¹₅B + ₊₁e⁰

Positron emission

2. ³⁵₁₆S → ? + ⁰₋₁e (beta decay so increase in atomic number)

³⁵₁₆S → ³⁵Cl17 + ⁰₋₁e

3. ? → ⁹⁰₃₉Y + ⁰₋₁e (beta decay atomic number would be increased in product)

So in reactant decrase atomic number by one

⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e

4.

²¹⁰₈₃Bi → ? + ⁴₂He

Equate the atomic number

83 = x +2

x = 83-2 = 81

equate the mass number

210 = x + 4

x = 210 - 4 = 206

²¹⁰₈₃Bi → ²⁰⁶Tl₈₁ + ⁴₂He

5

? --->  ⁸⁹Y₃₉ + ⁰₊₁e

No change in mass

So equate the atomic number

x = 39 + 1 = 40

⁸⁹Zr₄₀--->  ⁸⁹Y₃₉ + ⁰₊₁e