`.^(23)Na` is the more stable isotope of Na. Find out the process by which `._(11)^(24)Na` can undergo radioactive decayA. `beta^(-)` emssionB. `alpha`- emissionC. `beta^(+)` emssionD. K electron capture

`.^(23)Na` is the more stable isotope of Na. Find out the process by w

1.	`.^(23)Na` is the more stable isotope of Na. Find out the process by which `._(11)^(24)Na` can undergo radioactive decayA. `beta^(-)` emssionB. `alpha`- emissionC. `beta^(+)` emssionD. K electron capture
Answer» Correct Answer - A n/p ratio of `.^(24)Na` nuclide is 13/11 i.e., greater than unity and hence radioactive. To achieve stability, it would tend to adjust its n/p ratio to the proper value of unity. This can be done by breaking a neutron into proton and electron. `._(0)n^(1) rarr ._(+1)P^(1) + ._(-1)e^(0) or beta^(-)` The proton will stay inside the nucleus whereas electron which cannot exist in the nucleus, will be emitted out as `beta-` ray

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`.^(23)Na` is the more stable isotope of Na. Find out the process by which `._(11)^(24)Na` can undergo radioactive decayA. `beta^(-)` emssionB. `alpha`- emissionC. `beta^(+)` emssionD. K electron capture

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