
The activity of a radioactive substance denoted by A is equal to:
 `(dN)/dt = lamdaN`
 `(dN)/dt = lamdaN`
 `(dN)/dt = lamda/N`
 `(dN)/dt = N/lamda`
 N/A
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Answer: B
Explanation: 
The unit of activity is the curie (Ci) which is the rate of decay of:
 3.7 x 10^{10} dps
 3.7 x 10^{10} dps
 2.7 x 10^{11} dps
 6 x 10^{10} dps
 Where dps is disintegration or decay per second.
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Answer: A
Explanation: 
The S.I unit of activity is:
 Curie (Ci)
 Rutherford
 Becquerel (Bq)
 none of these
 The S.I unit of activity is Becquerel (Bq) and is equal to one disintegration per second (dps).
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Answer: C
Explanation: 
The activity of a radioactive substance is commonly measured by:
 Geigermuller counter
 Cloud chamber
 Film Badge
 Scintillation counter
 N/A
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Answer: A
Explanation: 
Au198 has a half life of 2.7 days. If you start with 10 mg of a sample of Au198, how much its amount will remain after 10 days?
 1.30 mg
 0.768 mg
 0.246 mg
 1.20 mg
 Less than four halflives almost 3.7 halflives. Thus amount left = `10/2^3.7` = 0.768
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Answer: B
Explanation: 
The halflife of P32 is 14.3 days. How many days are required to drop to 6.25% of its initial amount?
 42.9
 28.6
 50
 57.2
 6.25% amount will remain after 4 halflives. Thus days required = 14.3 x 4 = 57.2
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Answer: D
Explanation: 
The halflife for the beta decay of Pa233 is 27.4 days. How many days are required to reduce a 5.0 g sample of Pa233 to 0.625 g?
 82.2
 109.6
 54.8
 27.4
 After 3 halflives (27.4 x 3 = 82.2 days), the amount left will be = `(5g)/2^3=5/8` = 0.625 g,
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Answer: A
Explanation: 
How many halflives are required for a radioactive isotope to reduce to 12.5% of its initial amount?
 2
 3
 4
 5
 After three halflives, the %age amount left behind = `100/2^3` = 12.5%
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Answer: B
Explanation: 
If 93.75% amount of Iodine131 decays in 32 days. Its halflife perod is:
 10.6 days
 6.4 days
 5.3 days
 8 days
 93.75% amount decays after 4 halflives. After 4 halflives (4 x 8 =32 days), the %age amount left = `100/2^4` = 6.25% Thus decaying amount = 100  6.25 = 93.75%.
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Answer: D
Explanation: 
The halflife of tritium (H3) is 12.3 years. If 48.0 mg of it is releaed from a nuclear power plant during the course of an accident. What mass of this nuclide will remain after 49.2 years?
 6.0 mg
 24.0 mg
 3.0 mg
 12.0 mg
 After 4 halflives (12.3 x 4 = 49.2 years), the amount left will be = `(48mg)/2^4` = 3.0 mg
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Answer: C
Explanation: