Chlorine has two stable nuclides, \({ }^{35} \mathrm{Cl}\) and ${ }^{37} \mathrm{Cl}\(. In contrast, \){ }^{36} \mathrm{Cl}$ is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \({ }^{36} \mathrm{Cl}\) ? (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \({ }^{36} \mathrm{Cl}\) is less stable than either ${ }^{35} \mathrm{Cl}\( or \){ }^{37} \mathrm{Cl}$.

Short Answer

Expert verified
(a) The product of decay of \({ }^{36}\mathrm{Cl}\) is \({ }^{36}\mathrm{Ar}\). (b) \({ }^{36}\mathrm{Cl}\) is less stable than \({ }^{35}\mathrm{Cl}\) and \({ }^{37}\mathrm{Cl}\) primarily due to its odd N and odd Z combination, which is associated with increased instability in isotopes with odd atomic numbers.

Step by step solution

01

(a) Identify decay product of \({ }^{36}\mathrm{Cl}\) through beta emission

When a nuclide undergoes beta emission (β⁻ decay), a neutron in the nucleus is transformed into a proton. This increases the atomic number (number of protons) by 1, while the mass number (total of protons and neutrons) remains unchanged. The equation for the beta decay can be written as: \[n \rightarrow p^{+} + e^{-} + \overline{\nu}_e\] Understanding this process, let's find the decay product of \({ }^{36}\mathrm{Cl}\). The mass number will still be 36, and the atomic number will increase by 1. This atomic number of 18 corresponds to Argon (Ar). So, the product of the decay of \({ }^{36}\mathrm{Cl}\) is \({ }^{36}\mathrm{Ar}\).
02

(b) Compare the stability of \({ }^{36}\mathrm{Cl}\) to \({ }^{35}\mathrm{Cl}\) and \({ }^{37}\mathrm{Cl}\)

We will examine the neutron-to-proton ratios (N/Z) of \({ }^{35}\mathrm{Cl}\), \({ }^{36}\mathrm{Cl}\), and \({ }^{37}\mathrm{Cl}\), as well as whether they have any magic numbers contributing to their stability. 1. \({ }^{35}\mathrm{Cl}\): - Protons (Z) = 17 - Neutrons (N) = 35 - 17 = 18 - N/Z = 18/17 = 1.06 2. \({ }^{36}\mathrm{Cl}\): - Protons (Z) = 17 - Neutrons (N) = 36 - 17 = 19 - N/Z = 19/17 = 1.12 3. \({ }^{37}\mathrm{Cl}\): - Protons (Z) = 17 - Neutrons (N) = 37 - 17 = 20 - N/Z = 20/17 = 1.18 As we can see, \({ }^{36}\mathrm{Cl}\) has a lower N/Z ratio than \({ }^{37}\mathrm{Cl}\) but higher than \({ }^{35}\mathrm{Cl}\). However, while having stable isotopes with odd atomic numbers, nuclides with an odd N and odd Z are generally unstable, making \({ }^{36}\mathrm{Cl}\) less stable than \({ }^{35}\mathrm{Cl}\) and \({ }^{37}\mathrm{Cl}\). Examining for magic numbers, none of these nuclides have neutron or proton numbers associated with recognized magic numbers that organize stable shells. Thus, magic numbers don't play a role in the stability contrast between these Chlorine isotopes. In conclusion, the lower stability of \({ }^{36}\mathrm{Cl}\) as compared to \({ }^{35}\mathrm{Cl}\) and \({ }^{37}\mathrm{Cl}\) relies primarily on its odd N and odd Z combination, which is associated with increased instability in isotopes with odd atomic numbers.

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