Question

The end product formed in the disintegration of $^{222}_{88}$ Ra is:
A) $^{204}_{81}$ Tl
B) $^{206}_{82}$ Pb
C) $^{222}_{86}$ Rn
D) $^{207}_{83}$ Bi

Answer 1

Hint: The end product formation in the disintegration is based on the radioactive decay. The disintegration can occur through alpha decay, beta decay, and gamma decay. Identify the decay in Radium with the atomic number 88; and the end product could be known.

Complete step by step answer:
*First, let us discuss the types of radioactive decay i.e. alpha, beta, and gamma decay.
*First, we will know about the alpha decay. When a nucleus emits an alpha particle, in consideration with the nucleus of helium as an alpha particle. There are two protons, and two neutrons.
*If we talk about the beta particle, it is considered to be a positron, or an electron, there is an increase in proton number.
*Now, talking about the gamma decay, the photon will be emitted.
*If we see the disintegration of a given element $^{222}_{88}$ Ra, then it will show the alpha, and beta decay.
*It means the emission of helium nucleus, as well as an electron (beta particle).
*The reaction can be represented as:
$^{222}_{88}$ Ra $\rightarrow$ $^{206}_{82}$ Pb + 4 $^{4}_{2}$ He + 2 ($^{0}_{-1}$ $\beta$)
*So, we can say that this is the balanced disintegration equation of radium with the atomic number 88, and it leads to the formation of element lead with the atomic number 82.
*In the last, we can conclude that the correct option is (B).

Note: Don’t get confused while identifying the end product in this disintegration. Here, we have considered the both decays, because in consideration with the single decay, we cannot attain the elements in the given option. Thus, we attained the lead element in consideration with both the decays.