Question

$_{90}^{232}Th$ an isotope of thorium decays in ten stages emitting six a-particles and four b-particles in all. The end product of the decay is
A. $_{82}^{206}Pb$
B. $_{82}^{209}Pb$
C. $_{82}^{208}Pb$
D. $_{83}^{209}Br$

Answer 1

Hint: In the thorium isotope decay series, thorium loses 6 alpha particles and 4 beta particles in ten stages of the process. And alpha disintegration is the type of radioactive disintegration. It is explained where a radioactive nucleus emits an alpha particle that is 42He. In the nuclear term some element of beta decay could produce a daughter nucleus of the element adjacent to the periodic table.


Complete answer:
In this case of alpha and beta decay let's name it as B. and the element produced by the beta decay off the isotope of thorium-232 is a protactinium-232 and it is denoted by the symbol Pa. Thorium isotope has its natural origin in 232-thorium occurs with 100 percent of isotopic abundance. The gaseous product in Thorium disintegration is Thorium, it has a very short period of half-life as 56s.
For alpha decay reaction is written as below:
$_Z{X^A}{ \to _2}H{e^4}{ + _{Z - 1}}{Y^{A - 4}}$
In beta decay is the type of radioactive disintegration process that is where a parent nucleus emitted an electron. Some neutrons in the parent nucleus can split up into an electron and a neutron. If once the parent nucleus emits an electron, then there must be an additional proton increasing the atomic number of the nucleus. In the terms of the atomic nucleus of beta disintegration it can produce the daughter nucleus. And in this case it can express as below:
$_Z{X^A}{ \to _{ - 1}}{e^0}{ + _{Z + 1}}{Y^A}$
Where, Z is the atomic number of the element; A is the isotopic number.
The end product of disintegration is expressed as below:
${_{90}{Th^{232}}\xrightarrow{6\alpha}{_{Z^{'}}}{X^A}}$
This equation explains that the Thorium-232 decays with the emission of 6α particles, so we can get the end product of alpha decay.
A’ = A‒6 × 4
     = 232 ‒ 24
 A’ = 208
  Z’ = Z ‒ 2 × 6
  Z’ = 90‒ 12
 Z’ = 78
In beta disintegration the element increases by 4 because of the emission of four beta particles. So that we can get below :
The atomic number is increased by 4 due to the emission of beta particles.
Again, Z = 78 + 4. Here, we already know the atomic number value from above. So we can get the atomic number of the element is,
Z = 78 + 4 = 82
${_{90}{Th^{232}}\xrightarrow{6\alpha}{_{78}}{X^{208}}\xrightarrow{4\beta}{_{82}}{Pb^{208}}}$
Thus the end of the product of the alpha and beta disintegration is lead-208.
Hence, the correct answer is option C.




Note: In radioactive decay the number of protons is equal to the atomic number. The given atomic number of Thorium is 90 and we get the atomic number at the end of the product is Lead 82 due to the emission of 6 alpha particles and the emission of 4 beta particles.