Photo AI
Nuclear disintegrations occur in radioactivity and in fission - Leaving Cert Physics - Question 8 - 2005
Question 8
Nuclear disintegrations occur in radioactivity and in fission. Distinguish between radioactivity and fission. Give an application of (i) radioactivity, (ii) fissi... show full transcript
Worked Solution & Example Answer:Nuclear disintegrations occur in radioactivity and in fission - Leaving Cert Physics - Question 8 - 2005
Step 1
Distinguish between radioactivity and fission.
Answer
Radioactivity is the disintegration of an unstable nucleus with the emission of alpha (α), beta (β), or gamma (γ) radiation. It involves the spontaneous transformation of an atom.
Fission, on the other hand, involves the splitting of a large nucleus into two smaller, similar-sized nuclei, accompanied by the release of energy and neutrons.
Step 2
Give an application of (i) radioactivity, (ii) fission.
Answer
(i) Applications of radioactivity include carbon dating, medical applications such as cancer treatment and sterilization, and smoke detectors.
(ii) Applications of fission include generating electrical energy in nuclear power plants and its use in nuclear weapons.
Step 3
Step 4
Describe an experiment to demonstrate the ionising effect of radioactivity.
Answer
-
Apparatus: Use a radioactive source and a charged (gold leaf) electroscope.
-
Procedure: Bring the radioactive source close to the cap of the electroscope.
-
Observation: The electroscope's leaves will collapse as the ionising radiation from the source ionizes the air around it.
-
Conclusion: The electroscope's leaves collapse due to the ionization of air, indicating that the radioactive source releases ionizing radiation.
Step 5
Step 6
Calculate the decay constant of cobalt-60.
Answer
To calculate the decay constant (( \lambda \)), use the formula:
[ \lambda = \frac{\ln(2)}{T_{1/2}} ]
where ( T_{1/2} = 5.26 , \text{years} = 5.26 \times 10^6 , \text{s} ).
Substituting the values gives:
[ \lambda = \frac{\ln(2)}{5.26 \times 10^6 , \text{s}} = 4.18 \times 10^{-9} , \text{s}^{-1} ]
Step 7
Calculate the rate of decay of a sample of cobalt-60 when it has 2.5 × 10^21 atoms.
Answer
The activity (A) of a sample is given by the formula:
[ A = \lambda N ]
where ( \lambda = 4.18 \times 10^{-9} , \text{s}^{-1} ) and ( N = 2.5 \times 10^{21} ).
Calculating the activity gives:
[ A = 4.18 \times 10^{-9} , \text{s}^{-1} \times 2.5 \times 10^{21} = 10.45 \times 10^{12} , \text{Bq} ]
Thus, the rate of decay is approximately 10.45 trillion decays per second.