InterviewSolution
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InterviewSolution
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`._(11)^(23)Na` is the most stable isotope of Na. Find the process by which `._(11)^(24) Na` can undergo radioactive decay :A. `beta^(-)`-emissionB. `alpha`-emissionC. `beta^(+)`-emissionD. K-electron capture |
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Answer» Correct Answer - A Since `._(11)^(23)Na` is the stable isotope, in `._(11)Na^(24) ` n/p ratio is higher than required. It will decompose by `beta^(-)` emission , all the other three methods (`alpha` emission, `beta^(+)` emission, k-electron capture) increase n/p ratio. `{:(._(11)Na^(24),rarr,._(12)Mg^(24)+._(-1)e^(0),),(("unstable"),,("stable"),),(,,,):}` |
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