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:

  1. ${ } _{13} Al^{29}$ is neutron rich isotope, will decay by $\beta$-emission converting some of its neutron into proton as

$$ { } _0 n^{1} \longrightarrow{ } _{-1} \beta^{0}+{ } _1 H^{1} $$



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