The last two changes in a radioactive decay series are shown below - Scottish Highers Physics - Question 10 - 2016
Question 10
The last two changes in a radioactive decay series are shown below.
A Bismuth nucleus emits a beta particle and its product, a Polonium nucleus, emits an alpha part... show full transcript
Worked Solution & Example Answer:The last two changes in a radioactive decay series are shown below - Scottish Highers Physics - Question 10 - 2016
Step 1
Identify P: Bismuth nucleus (Bi)
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Answer
The Bismuth nucleus (Bi) has a mass number of 212 (as from the decay table) and atomic number 83. Therefore, P is 212.
Step 2
Determine Q: Beta decay product
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Answer
During beta decay, a neutron is transformed into a proton, increasing the atomic number by 1 while keeping the mass number the same. Thus, Q will be 212 with atomic number 84.
Step 3
Identify R: Polonium nucleus (Po)
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The Polonium nucleus (Po), which receives the product from the decay of Bismuth, will therefore have the mass number of 208 (same as before) and an atomic number of 84.
Step 4
Determine S: Alpha decay product
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Answer
In the alpha decay of Polonium, two protons and two neutrons are lost, reducing the mass number by 4 and the atomic number by 2. Therefore, S will be 204 for the mass number and 82 for the atomic number.
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