AQA A-Level Physics Year 2 - Radioactivity question
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I need help with the following question:
The energy source for a remote weather station is the radioactive isotope of strontium 90/38 Sr which has a half-life of 28 years. It emits beta particles of energy 0.40 MeV. For a mass of 10g of this isotope, calculate:
i) its activity
ii) the energy released per second
i) For this question I managed to successfully obtain 5.2 x 10^13 using the radioactivity equations
ii) For this question I think you need to use the energy of 1 beta particle at 0.4 MeV but I'm not sure on how to get the value to be per second or which equation(s) to use.
Help is much appreciated.
The energy source for a remote weather station is the radioactive isotope of strontium 90/38 Sr which has a half-life of 28 years. It emits beta particles of energy 0.40 MeV. For a mass of 10g of this isotope, calculate:
i) its activity
ii) the energy released per second
i) For this question I managed to successfully obtain 5.2 x 10^13 using the radioactivity equations
ii) For this question I think you need to use the energy of 1 beta particle at 0.4 MeV but I'm not sure on how to get the value to be per second or which equation(s) to use.
Help is much appreciated.
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#2
The activity multiplied by the energy of the beta particle will equal the energy emitted per second
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(Original post by alevelphysicist)
The activity multiplied by the energy of the beta particle will equal the energy emitted per second
The activity multiplied by the energy of the beta particle will equal the energy emitted per second

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(Original post by alevelphysicist)
The activity multiplied by the energy of the beta particle will equal the energy emitted per second
The activity multiplied by the energy of the beta particle will equal the energy emitted per second
The radioactive isotope of iodine, 131/53 I, is used for medical diagnosis of the kidneys. A sample of the isotope is to be given to a patient in a glass of water. The passage of the isotope through each kidney is then monitored using two detectors outside the body.
i) The reading from the detector near one of the patient's kidneys rises then falls. The reading from the other detector which is near the other kidney rises and does not fall. Discuss the conclusions that can be drawn from these observations.
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(Original post by 12westlada)
Do you have any ideas for this question?
The radioactive isotope of iodine, 131/53 I, is used for medical diagnosis of the kidneys. A sample of the isotope is to be given to a patient in a glass of water. The passage of the isotope through each kidney is then monitored using two detectors outside the body.
i) The reading from the detector near one of the patient's kidneys rises then falls. The reading from the other detector which is near the other kidney rises and does not fall. Discuss the conclusions that can be drawn from these observations.
Do you have any ideas for this question?
The radioactive isotope of iodine, 131/53 I, is used for medical diagnosis of the kidneys. A sample of the isotope is to be given to a patient in a glass of water. The passage of the isotope through each kidney is then monitored using two detectors outside the body.
i) The reading from the detector near one of the patient's kidneys rises then falls. The reading from the other detector which is near the other kidney rises and does not fall. Discuss the conclusions that can be drawn from these observations.
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(Original post by alevelphysicist)
Well in one kidney, the iodine entering the kidney and then passing through, which would probably indicate that this kidney was working correctly. Whereas in the other kidney, there is not a reduction and hence this kidney is not able to filter out the iodine after it enters the kidney and so is probably failing.
Well in one kidney, the iodine entering the kidney and then passing through, which would probably indicate that this kidney was working correctly. Whereas in the other kidney, there is not a reduction and hence this kidney is not able to filter out the iodine after it enters the kidney and so is probably failing.
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