Question

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

(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 \({ }^{37}\mathrm{Cl}\) because it has an odd number of neutrons and a neutron-to-proton ratio further away from 1. The empirical rules of nuclear stability favor nuclei with even numbers of protons and neutrons and a neutron-to-proton ratio close to 1 for lighter elements.

Step-by-step solution

Identify the decay process

Since we are told that \({ }^{36}\mathrm{Cl}\) decays by beta emission, we need to recall that beta decay is a process that involves the conversion of a neutron into a proton (or vice versa) within the nucleus. There are two types of beta decay: 1. Beta-minus decay (\(\beta^-\)): a neutron is converted into a proton, and an electron - called a beta particle - is emitted. 2. Beta-plus decay (\(\beta^+\)): a proton is converted into a neutron, and a positron - the antiparticle of an electron - is emitted.

Write the decay equation for \({ }^{36}\mathrm{Cl}\)

We are dealing with the decay of \({ }^{36}\mathrm{Cl}\), which has an atomic number (number of protons) of 17 and a mass number (number of protons + number of neutrons) of 36. We know that a neutron within the nucleus of \({ }^{36}\mathrm{Cl}\) is converted into a proton during beta-minus decay, so the atomic number will increase by 1, and the mass number will remain unchanged. The decay equation can be written as: \[{}_Z^A X \rightarrow {}_{Z+1}^A Y + ^0_{-1}e\] Where \(X\) represents the original nuclide, \(Y\) represents the product nuclide, and \(e\) represents the beta particle (electron) emitted during decay.

Determine the product of decay for \({ }^{36}\mathrm{Cl}\)

Plugging the values of \({ }^{36}\mathrm{Cl}\) into the decay equation, we have: \[{}_{17}^{36} \mathrm{Cl} \rightarrow {}_{18}^{36} Y + ^0_{-1}e\] Since the decay process results in the emission of a beta particle and an increase in atomic number by 1, the product element will be the one with atomic number 18, which is Argon. Therefore, the product of decay is \({ }^{36}\mathrm{Ar}\).

Analyze the empirical rules of nuclear stability

There are a few rules (empirical rules) that help us understand nuclear stability: 1. Even number of protons and neutrons: Nuclei with even numbers of both protons and neutrons are more stable than those with odd numbers. 2. Neutron-to-proton ratio (\(n/p\)): For lighter elements (atomic number Z < 20), the most stable nuclei have an \(n/p\) ratio close to 1.

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

Let's check the neutron and proton numbers for each isotope: 1. \({ }^{35}\mathrm{Cl}\): 17 protons, 18 neutrons: n/p ratio = 18/17 ≈ 1.06 2. \({ }^{37}\mathrm{Cl}\): 17 protons, 20 neutrons: n/p ratio = 20/17 ≈ 1.18 3. \({ }^{36}\mathrm{Cl}\): 17 protons, 19 neutrons: n/p ratio = 19/17 ≈ 1.12 Applying the empirical rules, \({ }^{35}\mathrm{Cl}\) and \({ }^{37}\mathrm{Cl}\) have an even number of neutrons, making them more stable than \({ }^{36}\mathrm{Cl}\), which has an odd number of neutrons. Moreover, the \(n/p\) ratio in \({ }^{36}\mathrm{Cl}\) is further away from 1 than \({ }^{35}\mathrm{Cl}\), also suggesting that \({ }^{36}\mathrm{Cl}\) is less stable than \({ }^{35}\mathrm{Cl}\). Therefore, the overall stability of Cl isotopes can be ranked as: \({ }^{35}\mathrm{Cl} > { }^{37}\mathrm{Cl} > { }^{36}\mathrm{Cl}\).