Question

When ${}_{90}T{h}^{238}$ changes into ${}_{83}B{i}^{222}$, then the number of emitted $\alpha$ and $\beta$ particles are ?

• A. $8\alpha ,7\beta$
• B. $4\alpha ,7\beta$
• C. $4\alpha ,4\beta$
• D. $4\alpha ,1\beta$

Answer 1

The correct option is D $4\alpha ,1\beta$

$\begin{array}{c}\text{balancing mass no.}\\ 238=222+4x+0\\ \begin{array}{c}4x=16\\ \text{which means}4\alpha \text{particle released}\end{array}\end{array}$

$\begin{array}{c}\text{balaneing atomie no.}\\ \begin{array}{cc}90& =83+2x+y(-1)\\ 7& =8-y\\ y& =1\\ \text{So,}& \text{1}\beta \text{particle released}\\ 4\alpha ,1\beta & \text{particle released}\end{array}\end{array}$