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)
Show Answer
Answer:
Correct Answer: 5. (b)
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} $
BETA
