Question

A radioactive reaction is ${}_{92}{U}_{238}{\to }_{82}P{b}^{206}$. How many $\alpha$ and $\beta$ particles are emitted?

• A. $10\alpha ,6\beta$
• B. $4proton,8neutron$
• C. $6electron,8proton$
• D. $6\beta ,8\alpha$

Answer 1

The correct option is D $6\beta ,8\alpha$

$\begin{array}{c}{\text{Given - * Radioactive Reaction}}_{92}^{238}\cup ⟶{}_{82}^{206}Pb+\text{Particles.}\\ \text{From the Reaction, we conclude that -}\\ \text{*}\text{loss of atomic mass in daughter nuclei}\\ =32\text{units.}\end{array}$

$\begin{array}{c}\text{* Loss of atomic number}=10\text{units.}\\ \text{we know that -}\\ \text{* one}\alpha \text{-particle decreases atomic}\\ \text{- mass by}4\text{units.}\\ \text{- number by}2\text{units.}\end{array}$

$\begin{array}{c}\text{* one}\beta \text{-particle increase the atomic}\\ \text{and bring no change in atomic mass.}\\ \Rightarrow \text{For change in atomic mass of}32\text{units}\\ \text{we can say}8\alpha \text{-particles are released .}\\ \text{as only}\alpha \text{-particle can change atomic mass.}\end{array}$

$\begin{array}{c}\Rightarrow 8a\text{- partides decrease the atomic no. by}\\ 16\text{units but the reaction specifies that only}\\ 10\text{units of change in atomic mass.}\\ \text{So}6\beta \text{particles are also released to compensate}\\ \text{this change.}\\ \text{Hence, option (d) is correct.}\end{array}$