Question

A radioactive element $90X^{238}$decay into$83Y^{222}$. The number of $\beta -$particles emitted are.

• A. 4
• B. 6
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Number of $\alpha -$particles emitted $= \frac{238 - 222}{4} = 4$

This decreases atomic number to $90 - 4 \times 2 = 82$

Since atomic number of $83Y^{222}$is 83, this is possible if one $\beta -$particle is emitted.