1.

Which is more unstable of each of the following pairs, and in each case what type of process could the unstable nucleus undergo? a. `._(6)C^(16)` b. `._(9)F^(18)`, `._(10)Ne^(18)`A. a. `._(6)C^(16)` ,`._(7)N^(16)`B. b. `._(9)F^(18)` `._(10)Ne^(18)`C.D.

Answer» In this case although `._(6)C^(16)` has even number of `p` and `n`, still its `n//p` ratio is higher than `n//p` ratio of `._(7)N^(16)`,
`(n)/(p)` of `._(6)C^(16) = (10)/(6) = 1.7` `({:("For" Z = 20 n//p),("should be" = 1),("Forstability"):})`
`n//p` ratio is higher than 1. IT would emit `beta^(ɵ)` to get `n//p` ratio back to stable range.
`._(6)C^(16) rarr ._(7)N^(14) + ._(-1)e^(0)`
`(n)/(p) = (10)/(6) = 1.7, (n)/(p) = (7)/(7) = 1`
In case of `._(7)N^(16), (n)/(p) = (9)/(7) = 1.2`
`n//p` of `._(6)C^(16) gt n//p` of `._(7)N^(16)`. So `_(6)C^(16)` more unstalbe than `._(7)N^(16)`.
b. In this case although `._(9)F^(18)` has odd number `p` and `n`, still its `n//p` ratio is equal 1, has `n//p` of `._(10)Ne^(18)` is less than 1.
`(n)/(p)` of `._(9)F^(18) = (9)/(9) = 1` (For `Z = 20`, stability `n//p = 1)`
`(n)/(p)` of `._(10)Ne^(18) = (8)/(10) = 0.8` (Its value is less than 1)
So `._(10)Ne^(18)` would be unstable than `._(9)F^(18)`.
So, it will emitt `beta^(o+)` or K-capture to get the ratio into the range of stability.
`._(10)Ne^(18) rarr ._(9)F^(18) + ._(+1)e^(0)`
`(n)/(p) = (8)/(10) = 0.8` or `(n)/(p) = (9)/(9) = 1 (0.8 rarr 1)`
`._(10)Ne^(18) + ._(+1)F^(18)`


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