Radioactive Series

Q.

Thorium disintegration series is known as:

1. (4n + 2) series
2. (4n + 3) series
3. (4n + 1) series
4. (4n) series

Q. The end product of $(4n+2)$ radioactive disintegration series is

1. ${}_{82}P{b}^{207}$
2. ${}_{83}B{i}^{210}$
3. ${}_{82}P{b}^{208}$
4. ${}_{82}P{b}^{206}$

Q. In the given radioactive disintegration series,
${}_{90}^{232}Th{\to }_{82}^{208}Pb$
Calculate the value of (n+2) where value of n is number of isobars formed in this series emission of $\beta$ − particles.

Q. The element ${}_{90}^{232}Th$ belongs to 4n series which will be the end product of this series:

1. ${}_{83}^{209}Bi$
2. ${}_{82}^{208}Bi$
3. ${}_{82}^{209}Bi$
4. ${}_{82}^{207}Bi$

Q. The end product of $(4n+2)$ radioactive disintegration series is

1. ${}_{82}P{b}^{208}$
2. ${}_{82}P{b}^{206}$
3. ${}_{82}P{b}^{207}$
4. ${}_{83}B{i}^{210}$

Q. The end product of $(4n+2)$ radioactive disintegration series is

1. ${}_{82}P{b}^{208}$
2. ${}_{82}P{b}^{206}$
3. ${}_{82}P{b}^{207}$
4. ${}_{83}B{i}^{210}$