Question

# When $$_{90}Th^{238}$$ changes into $$_{83}Bi^{222}$$, then the number of emitted $$\alpha$$ and $$\beta$$ particles are ?

A
8α,7β
B
4α,7β
C
4α,4β
D
4α,1β

Solution

## The correct option is D $$4\alpha,1\beta$$$$\begin{array}{l}\text { balancing mass no. } \\\qquad 238=222+4 x+0 \\\qquad \begin{array}{l}4 x=16 \\\text { which means } 4 \alpha \text { particle released }\end{array}\end{array}$$\begin{array}{l}\text { balaneing atomie no. } \\\qquad \begin{aligned}90 &=83+2 x+y(-1) \\7 &=8-y \\y &=1 \\\text { So, } & \text { 1 } \beta \text { particle released } \\4 \alpha, 1 \beta & \text { particle released }\end{aligned}\end{array}Physics

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