Question

A radioactive element $_{90}X^{238}$ decay into $_{83}Y^{222}$ . The no of $\beta$ particles emitted are

• A. 4
• B. 6
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Answer 2

No. of $\alpha$ - particles emitted = $\frac{238 - 222}{4}$ = 4 This decreases the atomic number to 90 - 4 $\times$ 2 = 82 Since atomic number of $_{83}Y^{222}$ is 83, this is possible if one $\beta$-particle is emitted.