Question

By the disintegration of\[{}_{94}P{u^{241}}\], the element which is produced is also radioactive and disintegrates. In such a series total \[8\alpha \]-particles and \[5\beta \]-particles are emitted and then the process stops. Which is the final element produced?
\[(A){}_{83}B{i^{209}}\]
\[(B){}_{82}P{b^{209}}\]
\[(C){}_{83}B{i^{214}}\]
\[(D){}_{82}P{b^{214}}\]

Answer 1

Hint : The end product formation within the disintegration is based on radioactive decay. The disintegration may occur through alpha decay, beta decay, and gamma decay. Identify the decay in Plutonium with the atomic number\[94\] ; and also the end product could be known.

Complete step-by-step solution:
First, let us discuss the categories of radioactive decay i.e. alpha, beta, and gamma decay.
First, we’ll discuss 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 therefore two neutrons.
If we discuss the \[\beta \] particle, it’s considered to be a positron, or an electron, there’s an increase in proton number.
Now, discussing the \[\gamma \] decay, the photon is going to be emitted.
If we see the disintegration of a given element\[{}_{94}P{u^{241}}\], then it’ll show the alpha and beta decay.
It means the emission of a helium nucleus, moreover as an electron (beta particle).
The reaction can be represented as:
\[{}_{94}P{u^{241}}\xrightarrow{{8\alpha }}{}_{78}{X^{209}}\xrightarrow{{5\beta }}{}_{83}{Y^{209}}\]
\[{}_{94}P{u^{241}} \to {}_{83}B{i^{209}} + 4{}_2H{e^4} + 2{}_{ - 1}{\beta ^0}\]
The mass number decreases by\[8 \times 4 = 32\].
Atomic number decreases by\[\left( {8 \times 2} \right) - 5 = 11\].
So, we’ll say that this is often the balanced disintegration equation of plutonium with the atomic number\[94\], and it leads to the formation of elements lead with the atomic number\[83\].
Hence, the correct option is A.

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