Interpretation:
Balanced
Concept Introduction:
Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.
Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.
- The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.
- The sum of
atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.
Explanation of Solution
In the decay series of 232Th , the sequence of particles emitted in each step follows as,
α, β−, β−, α, α, α, α, β−, β−, α
According to the law of conservation of
The
- Step (1): Emission of α particle from 232Th
An alpha particle is emitted and an alpha particle is 42α. Atomic number of Thorium is 90.
The given unbalanced nuclear equation is,
23290Th→ ? + 42α
Product formed by α decaying of 232Th is determined as follows,
Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product23290Th = x82Ra + 4(alpha)x = 228
The elemental symbol of the unknown element is 22888Ra. Therefore, the balanced nuclear equation is,
23290Th→ 22888Ra + 42α_
- Step (2): Emission of β− particle from 22888Ra
A β− particle is emitted and a β− particle is 0−1β.
The given unbalanced nuclear equation is,
22888Ra→ 0−1β + ?
Product formed by β− decaying of 22888Ra is determined as follows,
Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product22888Ra = x89Ac + 0(β−)x = 228
The elemental symbol of the unknown element is 22889Ac. Therefore, the balanced nuclear equation is,
22888Ra→ 0−1β + 22889Ac_
- Step (3): Emission of β− particle from 22889Ac
The given unbalanced nuclear equation is,
22889Ac → 0−1β + ?
Product formed by β− decaying of 22889Ac is determined as follows,
Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product22889Ac = x90Th + 0(β−)x = 228
The elemental symbol of the unknown element is 22890Th. Therefore, the balanced nuclear equation is,
22889Ac → 0−1β + 22890Th_
- Step (4): Emission of α particle from 22890Th
The given unbalanced nuclear equation is,
22890Th→ ? + 42α
Product formed by α decaying of 22890Th is determined as follows,
Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product22890Th = x88Ra + 4(alpha)x = 224
The elemental symbol of the unknown element is 22488Ra. Therefore, the balanced nuclear equation is,
22890Th→ 22488Ra+ 42α_
- Step (5): Emission of α particle from 22488Ra
The given unbalanced nuclear equation is,
22488Ra→ ? + 42α
Product formed by α decaying of 22488Ra is determined as follows,
Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product22488Ra = x86Rn + 4(alpha)x = 220
The elemental symbol of the unknown element is 22086Rn. Therefore, the balanced nuclear equation is,
22488Ra→ 22086Rn + 42α_
- Step (6): Emission of α particle from 22086Rn
The given unbalanced nuclear equation is,
22086Rn→ ? + 42α
Product formed by α decaying of 22086Rn is determined as follows,
Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product22086Rn = x84Po + 4(alpha)x = 216
The elemental symbol of the unknown element is 21684Po. Therefore, the balanced nuclear equation is,
22086Rn→ 21684Po + 42α_
- Step (7): Emission of α particle from 21684Po
The given unbalanced nuclear equation is,
21684Po→ ? + 42α
Product formed by α decaying of 21684Po is determined as follows,
Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product21684Po = x82Pb + 4(alpha)x = 212
The elemental symbol of the unknown element is 21282Pb. Therefore, the balanced nuclear equation is,
21684Po→ 21282Pb + 42α_
- Step (8): Emission of β− particle from 21282Pb
The given unbalanced nuclear equation is,
21282Pb→ 0−1β + ?
Product formed by β− decaying of 21282Pb is determined as follows,
Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product21282Pb = x83Bi + 0(β−)x = 212
The elemental symbol of the unknown element is 21283Bi. Therefore, the balanced nuclear equation is,
20982Pb → 0−1β + 21283Bi_
- Step (9): Emission of β− particle from 21283Bi
The given unbalanced nuclear equation is,
21283Bi → 0−1β + ?
Product formed by β− decaying of 21283Bi is determined as follows,
Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product21283Bi = x84Po + 0(β−)x = 212
The elemental symbol of the unknown element is 21284Po. Therefore, the balanced nuclear equation is,
21283Bi → 0−1β + 21284Po_
- Step (10): Emission of α particle from 21284Po
The given unbalanced nuclear equation is,
21284Po→ ? + 42α
Product formed by α decaying of 21284Po is determined as follows,
Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product21284Po = x82Pb + 4(alpha)x = 208
The elemental symbol of the unknown element is 20882Pb. Therefore, the balanced nuclear equation is,
21284Po→ 20882Pb + 42α_
Therefore, balanced nuclear equations for decay series of thorium-232 follows as,
23290Th→ 22888Ra + 42α22888Ra→ 0−1β + 22889Ac22889Ac → 0−1β + 22890Th22890Th→ 22488Ra+ 42α22488Ra→ 22086Rn + 42α22086Rn→ 21684Po + 42α21684Po→ 21282Pb + 42α20982Pb → 0−1β + 21283Bi21283Bi → 0−1β + 21284Po21284Po→ 20882Pb + 42α
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Chapter 24 Solutions
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