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Beta Decay and Radioactive Processes
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A radioactive element ${X} _{90}^{238}$ decays into ${Y} _{83}^{222}$. The number of $\beta$-particles emitted are
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A
1
💡 Explanation:
The radioactive element undergoes four alpha decays and one beta decay as follows:
$^{238} _{90}X\rightarrow ^{222} _{83}Y+4^4 _2He+ ^{0} _{-1}e$
Clearly only one beta particle (electron) is emitted.