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.(13)Al^(27) is a stable isotope. .(13)A...

`._(13)Al^(27)` is a stable isotope. `._(13)Al^(29)` is expected to disintegrate by

A

`alpha` emission

B

`_(-1)^(0) beta` emission

C

Positron emission

D

Proton emission

Text Solution

Verified by Experts

The correct Answer is:
B
Since `._(13)^(27) Al` is stable isotope , `._(13)^(29)Al` has `(n)/(P)` ratio higher than required, it will disintegrate by `beta^(-)` emission and `(n)/(P)` ratio will decrease.
`._(13)^(29)Al rarr ._(14) si^(29)+._(-1)e^(0)`
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Knowledge Check

  • ._(13)^(27)Al is a stable isotope. ._(13)^(29)Al is expected to disintegrate by

    A
    `alpha-` emission
    B
    `beta-emission
    C
    Positron emission
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    positron emission
    D
    proton emission

  • Unstable nuclei attain stability through disintergration. The nuclear stability is related to neutron- proton ratio (n//p) . n//p ratio for stable nuclei lies quite close to unity for elements with low atomic numbers ( 20 or less) but it is more than one for nuclei having higher atomic numbers. Nuclei having n//p ratio either very high or low undergo nuclear transformation. (i) When n//p ratio is higher than required for stability, the nuclei have the tendency to emit beta -rays, i.e. a neutron is converted into a proton. (ii) When n//p ratio is lower than required for stability, the nuclei increase the ratio, either by emitting alpha -particle or by emitting a positron or by K -electron capture. ""_(13)^(27)Al is a stable isotope , ""_(13)^(29)Al is expected to disintegrate by-

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    `alpha`-emission
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    D
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