in the 232, 90 Th decay series there are 6 radioisotopes that decay by alpha emission, including Th-232 itself

Question

and 4 radioisotopes that decay by B-. the final product is stable isotope. the symbol for this product is
A. 204/74 W
B. 208/74 W
C. 204/82Pb
D. 208/82 Pb
E. 208/78Pt

Answer 1

Georock is correct, but did not tell you how to approach solving this problem.

Think of it this way: each alpha decay changes the atomic number by -2 (each alpha particle takes away 2 protons), and changes the mass number by -4 (each alpha particle takes away 2 protons + 2 neutrons); each beta decay changes the atomic number by +1 (each decay converts a neutron into a proton) and does not affect the mass number.

We therefore have that:
final mass number = 232 - 6*4 = 208
final atomic number = 90 - 6*2 + 4*1 = 82

Answer D, 208/82Pb is the correct one.

Answer 2

You should be able to eliminate three of the answers very easily - W and Pt are not part of any radioactive decay series.

The answer is D - Pb 208, a stable element.