Question

A radioactive element $X_{90}^{238}$ decays into $Y_{83}^{222}$. The number of $\beta$-particles emitted are

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

The correct option is A 1

The radioactive element undergoes four alpha decays and one beta decay as follows:

${}_{90}^{238}X{\to }_{83}^{222}Y+4_2^{4}He{+}_{-1}^{0}e$

Clearly only one beta particle (electron) is emitted.