Pl 5. the following equations represent part of a radioactive decay series: 1. 92238u -> a + 24he 2. a -> 91234pa + b 3. 91234pa -> c + –10e 4. c -> 90230th + 24he 5. 90230th -> d + 24he 6. d -> 86222rn + e 7. 86222rn -> 84218po + f 8. 84218po -> g + 24he 9. g -> 83214bi + h 10. 83214bi -> i + –10e 11. i -> j + 24he fill in the table with the correct particle or element, and describe the types of radiation given off in the series.

Following is the balanced radioactive decay series:

Particle/radiations generated during the reaction are labeled in bold at end of reaction. 

Care must be taken that, atomic number and atomic mass number should be balanced in each of these reactions.

1) 92 238U →  90 234Th + 2 4He(α particle)

A = 90 234Th because alpha particle is emitted along with it. So atomic number of daughter element has to be 92 - 2 = 90. This corresponds to Th. 

2) 90 234Th  → 91 234Pa + -1 0e (electron)

B = -1 0e i.e electron because after radioactive disintegration atomic number of daughter element (Pa) is +1 as compared to parent element (Th)

3) 91 234Pa  → 92 234U + –1 0e (electron)

C = 92 234U because electron is emitted along with it. So atomic number of daughter element has to be 91 - (-1) = 92. This corresponds to U. 

4) 92 234U → 90 230Th + 2 4He (α particle)

In this case, 92 234U undergoes nuclear disintegration to generate 90 230Th and alpha particle

5) 90 230Th → 88 226Ra + 2 4He (α particle)

D = 88 226Ra because alpha particle is emitted along with it. So atomic number of daughter element has to be 90 - 2 = 88. This corresponds to Ra. 

6) 88 266Ra → 86 222Rn + 2 4He (α particle)

E = alpha particle because during nuclear disintegration, 88 266Ra is converted into 86 222Rn. Hence, for mass balance we have 88 - 86 = 2. It corresponds to alpha particles.

7) 86 222Rn → 84 218Po + 2 4He (α particle)

Again, F = alpha particle because during nuclear disintegration, 86 222Rn is converted into 84 218Rn. Hence, for mass balance we have 86 - 84 = 2. It corresponds to alpha particles.

8) 84 218Po → 82 214Pb + 2 4He (α particle)

G = 82 214Pb because alpha particle is emitted along with it. So atomic number of daughter element has to be 84 - 2 = 82. This corresponds to Pb.

9) 82 214Pb → 83 214Bi + -1 0e (electron)

H = -1 0e because after radioactive disintegration atomic number of daughter element (Bi) is +1 as compared to parent element (Pb)

10) 83 214Bi → 84 214Po + –1 0e (electron)

I = 84 214Po because electron is emitted along with it. So atomic number of daughter element has to be 83 - (-1) = 84. This corresponds to Po.

11) 84 214Po → 82 210Pb + 2  4He (α particle)

J = 82 210Pb because alpha particle is emitted along with it. So atomic number of daughter element has to be 84 - 2 = 82. This corresponds to Pb.

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