Nuclear Chemistry - Result Question 5
####5. ${ } _{13}^{27} Al$ is a stable isotope. ${ } _{13}^{29} Al$ is expected to decay by
(a) $\alpha$-emission
(b) $\beta$-emission
(c) positron emission
(d) proton emission
(1996, 1M)
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Answer:
Correct Answer: 5. (a)
Solution:
- ${ } _{13} Al^{29}$ is neutron rich isotope, will decay by $\beta$-emission converting some of its neutron into proton as
$$ { } _0 n^{1} \longrightarrow{ } _{-1} \beta^{0}+{ } _1 H^{1} $$
