Henri Becquerel was the first person to discover evidence of radioactivity - Leaving Cert Physics - Question d - 2022
Question d
Henri Becquerel was the first person to discover evidence of radioactivity. Radioactivity is the emission of radiation as a result of the decay of atomic nuclei.
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Worked Solution & Example Answer:Henri Becquerel was the first person to discover evidence of radioactivity - Leaving Cert Physics - Question d - 2022
Step 1
Name the other two types of radiation.
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Answer
The other two types of radiation are beta (β) and gamma (γ) radiation.
Step 2
Describe an experiment to show that the three types of radiation have different penetrating powers.
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Answer
To demonstrate the different penetrating powers of alpha, beta, and gamma radiation, you can set up the following experiment:
Apparatus:
Source of radiation: Use a sample containing alpha, beta, and gamma emitters, such as a radioactive isotope.
Barriers: Prepare different materials for testing penetration, such as paper (for alpha), aluminum sheet (for beta), and lead block (for gamma).
Method:
Place the radioactive source at a fixed distance from a Geiger counter or a radiation detector.
Measure the count rate of radiation detected without any barriers.
Place the paper barrier in front of the source and measure the count rate again.
Repeat this step with aluminum and lead barriers.
Observation/Detector:
Record how the count rate changes as each barrier is introduced. You will find that the count rate drops significantly with paper (for alpha), moderately with aluminum (for beta), and hardly with lead (for gamma), demonstrating the differences in penetrating power.
Step 3
How many neutrons are there in an atom of Ra^{226}?
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Answer
To find the number of neutrons in an atom of Ra^{226}, use the formula:
Number of neutrons = Mass number - Atomic number.
For radium (Ra), the atomic number is 88. Therefore:
Number of neutrons = 226 - 88 = 138.
Step 4
What is the daughter nucleus when an atom of Ra^{226} emits two alpha particles?
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Answer
When an atom of Ra^{226} emits two alpha particles, it loses a total of 4 nucleons (2 protons and 2 neutrons). The resultant daughter nucleus will be:
Mass number = 226 - 4 = 222
Atomic number = 88 - 2 = 86
Thus, the daughter nucleus is Polonium (Po^{222}).
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