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Q1(b) Write down the decay equations for the following radionuclides.

[tex]\[{}_{22}^{11} \text{Na} (\beta), {}_{27}^{27} \text{Co} (\text{EC}), {}^{127}_{55} \text{Cs} (\beta), {}^{241}_{95} \text{Am} (\alpha), {}^{252}_{98} \text{Cf} (\alpha)\][/tex]



Answer :

GinnyAnswer
Sure, let's write down the decay equations for the given radionuclides step-by-step:

1. Sodium-22 ([tex]\(_{22}^{11}\text{Na}\)[/tex]) undergoing beta decay ([tex]\(\beta^+\)[/tex]):

Sodium-22 ([tex]\(_{22}^{11}\text{Na}\)[/tex]) decays by emitting a positron ([tex]\(e^+\)[/tex]) and a neutrino ([tex]\(\nu_e\)[/tex]). The resulting daughter nucleus is Neon-22 ([tex]\(_{22}^{10}\text{Ne}\)[/tex]).

The decay equation is:
[tex]\[ { }_{22}^{11}\text{Na} \rightarrow { }_{22}^{10}\text{Ne} + e^+ + \nu_e \][/tex]

2. Cobalt-27 ([tex]\(_{27}^{27}\text{Co}\)[/tex]) undergoing electron capture (EC):

Cobalt-27 ([tex]\(_{27}^{27}\text{Co}\)[/tex]) captures an electron ([tex]\(e^-\)[/tex]), merging with a proton to form a neutron and converting into Iron-27 ([tex]\(_{27}^{26}\text{Fe}\)[/tex]). The process emits a neutrino ([tex]\(\nu_e\)[/tex]).

The decay equation is:
[tex]\[ { }_{27}^{27}\text{Co} + e^- \rightarrow { }_{27}^{26}\text{Fe} + \nu_e \][/tex]

3. Cesium-127 ([tex]\(_{127}^{55}\text{Cs}\)[/tex]) undergoing beta decay ([tex]\(\beta^-\)[/tex]):

Cesium-127 ([tex]\(_{127}^{55}\text{Cs}\)[/tex]) decays by emitting an electron ([tex]\(e^-\)[/tex]) and an antineutrino ([tex]\(\bar{\nu}_e\)[/tex]). The resulting daughter nucleus is Barium-127 ([tex]\(_{127}^{56}\text{Ba}\)[/tex]).

The decay equation is:
[tex]\[ { }_{127}^{55}\text{Cs} \rightarrow { }_{127}^{56}\text{Ba} + e^- + \bar{\nu}_e \][/tex]

4. Americium-241 ([tex]\(_{241}^{95}\text{Am}\)[/tex]) undergoing alpha decay ([tex]\(\alpha\)[/tex]):

Americium-241 ([tex]\(_{241}^{95}\text{Am}\)[/tex]) decays by emitting an alpha particle ([tex]\(_{4}^{2}\text{He}\)[/tex]). The resulting daughter nucleus is Neptunium-237 ([tex]\(_{237}^{93}\text{Np}\)[/tex]).

The decay equation is:
[tex]\[ { }_{241}^{95}\text{Am} \rightarrow { }_{237}^{93}\text{Np} + { }_{4}^{2}\text{He} \][/tex]

5. Californium-252 ([tex]\(_{252}^{98}\text{Cf}\)[/tex]) undergoing alpha decay ([tex]\(\alpha\)[/tex]):

Californium-252 ([tex]\(_{252}^{98}\text{Cf}\)[/tex]) decays by emitting an alpha particle ([tex]\(_{4}^{2}\text{He}\)[/tex]). The resulting daughter nucleus is Curium-248 ([tex]\(_{248}^{96}\text{Cm}\)[/tex]).

The decay equation is:
[tex]\[ { }_{252}^{98}\text{Cf} \rightarrow { }_{248}^{96}\text{Cm} + { }_{4}^{2}\text{He} \][/tex]

These are the detailed decay equations for the given radionuclides.

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