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`._(11)Na^(28)` is a radioactive and it decays toA. `._(12)Mg^(24)` and `beta`-particlesB. `._(11)Na^(21)` and neutronC. `._(13)F^(24)` and positronD. `._(9)F^(20)` and `alpha`-particles |
Answer» Correct Answer - a `._(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)` |
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