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A radioactive element ${ }_{90} X^{238}$ decays into ${ }_{83} Y^{222}$. The number of $\beta$-particles emitted are
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$\alpha$-particles are emitted $=\frac{238-222}{4}=4$
The atomic number is decreased
$90-4 \times 2=82$
As atomic number of ${ }_{83} Y^{222}$, So, atomic number is increased by 1 , therefore, one $\beta$-particle is emitted.
The atomic number is decreased
$90-4 \times 2=82$
As atomic number of ${ }_{83} Y^{222}$, So, atomic number is increased by 1 , therefore, one $\beta$-particle is emitted.
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