The stable nuclide ²⁰⁶₈₂Pb is formed from ²³⁸₉₂U by a long series of α and β⁻ decays. Which of the following nuclides could NOT
be involved in this decay series?
A) Th-234
B) U-239
C) Po-218
D) Bi-214
E) Rn-222
2 answers:
Answer:
Rn-222
Step-by-step explanation:
this decay takes specified step therefore Rn-222is not a decaying process
Answer:
B) U-239 is the answer
hope it helps you
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Answer:
70 decreases each stage
Step-by-step explanation:
350 / 5 = 70 decreases each stage
Since all decreases are equal and in 5 stages:
decreases * stages = total decreases
70 * 5 = 350
It would be (12,20) or (20,12) I don't know which one
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(g/f)(x)=g(x)/f(x)
Answer:
0 Jasmine did not insert her decimal point correctly after multiplying the actual amount
0 Jasmine’s estimate and actual product don’t match closely enough.
720 by 600! hope this helps
