The stable nuclide ²⁰⁶₈₂Pb is formed from ²³⁸₉₂U by a long series of α and β⁻ decays. Which of the following nuclides could NOT

Question

3 years ago

9

be involved in this decay series?
A) Th-234
B) U-239
C) Po-218
D) Bi-214
E) Rn-222

2 answers:

Makovka662 [10] · Answer 1 · 2022-09-16T11:54:59+03:00

8 0

Answer:

Rn-222

Step-by-step explanation:

this decay takes specified step therefore Rn-222is not a decaying process

Lana71 [14] · Answer 2 · 2022-09-16T14:06:39+03:00

6 0

Answer:

B) U-239 is the answer

hope it helps you