The radioactive nuclide $^{232}_{90}Th$ decays by one $\alpha$ emission followed by two $\beta^-$ emissions - AQA - A-Level Physics - Question 10 - 2020 - Paper 1
Question 10
The radioactive nuclide $^{232}_{90}Th$ decays by one $\alpha$ emission followed by two $\beta^-$ emissions.
Which nuclide is formed as a result of these decays?
A... show full transcript
Worked Solution & Example Answer:The radioactive nuclide $^{232}_{90}Th$ decays by one $\alpha$ emission followed by two $\beta^-$ emissions - AQA - A-Level Physics - Question 10 - 2020 - Paper 1
Step 1
Decay by one $\alpha$ emission
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Answer
When the nuclide 90232Th undergoes one α emission, it emits an α particle (which consists of 2 protons and 2 neutrons). This results in a decrease of 2 in the atomic number and a decrease of 4 in the mass number:
90232Th→88228Rn+α
Thus, the first nuclide formed is 88228Rn.
Step 2
Followed by two $\beta^-$ emissions
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Answer
In a β− decay, a neutron is converted into a proton, resulting in an increase of 1 in the atomic number while the mass number remains unchanged. Performing this twice on 88228Rn gives:
First β− emission:
88228Rn→89228Fr+β−
Second β− emission:
89228Fr→90228Th+β−
Thus, after two β− emissions, the final nuclide formed is 90228Th.
