`._(11)Na^(28)` is a radioactive and it decays to

Options

A. `._(12)Mg^(24)` and `beta`-particles
B. `._(11)Na^(21)` and neutron
C. `._(13)F^(24)` and positron
D. `._(9)F^(20)` and `alpha`-particles

Answer

a

Explanation

`._(11)Na^(24)` is a radioactive isotope and it decays to `._(12)Mg^(24)` and `beta`-particles, which is shown as
`._(11)Na^(24) to ._(12)Mg^(24) + ._(-1)beta^(0)`