PLSSSSS HEEEELP!!!!
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.

PLSSSSS HEEEELP 5 The following equations represent part of a radioactive decay series 1 92238U gt A 24He 2 A gt 91234Pa B 3 91234Pa gt C 10e 4 C gt 90230Th 24H class=

Respuesta :

PBCHEM
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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