Question

A radioactive element $_{90}{{X}^{238}}$ decays into $_{83}{{\Upsilon }^{222}}.$ The number of P-particle emitted are:

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

Answer: (A)

1

Answer 2

$\alpha$ -particle 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.