Q94 / 160 · JIPMER 2013 · Physics

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

  • Correct answer
    A
    1
  • B
    2
  • C
    4
  • D
    6

Solution

$\alpha$-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.

Hence, the correct answer is option: 'A'.

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