1.

.^(23)Na is the more stable isotope of Na. Find out the process by which ._(11)^(24)Na can undergo radioactive decay

Answer»

`beta^(-)` emssion
`alpha`- emission
`beta^(+)` emssion
K ELECTRON capture

Solution :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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