The activity of a radioactive substance denoted by A is equal to:

1. `-(dN)/dt = lamdaN`
2. `(dN)/dt = lamdaN`
3. `(dN)/dt = lamda/N`
4. `(dN)/dt = N/lamda`
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Answer: B
Explanation:
• N/A
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The unit of activity is the curie (Ci) which is the rate of decay of:

1. 3.7 x 10¹⁰ dps
2. 3.7 x 10^-10 dps
3. 2.7 x 10^-11 dps
4. 6 x 10¹⁰ dps
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Answer: A
Explanation:
• Where dps is disintegration or decay per second.
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The S.I unit of activity is:

1. Curie (Ci)
2. Rutherford
3. Becquerel (Bq)
4. none of these
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Answer: C
Explanation:
• The S.I unit of activity is Becquerel (Bq) and is equal to one disintegration per second (dps).
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The activity of a radioactive substance is commonly measured by:

1. Geiger-muller counter
2. Cloud chamber
3. Film Badge
4. Scintillation counter
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Answer: A
Explanation:
• N/A
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Au-198 has a half life of 2.7 days. If you start with 10 mg of a sample of Au-198, how much its amount will remain after 10 days?

1. 1.30 mg
2. 0.768 mg
3. 0.246 mg
4. 1.20 mg
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Answer: B
Explanation:
• Less than four half-lives almost 3.7 half-lives. Thus amount left = `10/2^3.7` = 0.768
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The half-life of P-32 is 14.3 days. How many days are required to drop to 6.25% of its initial amount?

1. 42.9
2. 28.6
3. 50
4. 57.2
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Answer: D
Explanation:
• 6.25% amount will remain after 4 half-lives. Thus days required = 14.3 x 4 = 57.2
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The half-life for the beta decay of Pa-233 is 27.4 days. How many days are required to reduce a 5.0 g sample of Pa-233 to 0.625 g?

1. 82.2
2. 109.6
3. 54.8
4. 27.4
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Answer: A
Explanation:
• After 3 half-lives (27.4 x 3 = 82.2 days), the amount left will be = `(5g)/2^3=5/8` = 0.625 g,
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How many half-lives are required for a radioactive isotope to reduce to 12.5% of its initial amount?

1. 2
2. 3
3. 4
4. 5
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Answer: B
Explanation:
• After three half-lives, the %age amount left behind = `100/2^3` = 12.5%
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If 93.75% amount of Iodine-131 decays in 32 days. Its half-life perod is:

1. 10.6 days
2. 6.4 days
3. 5.3 days
4. 8 days
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Answer: D
Explanation:
• 93.75% amount decays after 4 half-lives. After 4 half-lives (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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The half-life of tritium (H-3) 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?

1. 6.0 mg
2. 24.0 mg
3. 3.0 mg
4. 12.0 mg
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Answer: C
Explanation:
• After 4 half-lives (12.3 x 4 = 49.2 years), the amount left will be = `(48mg)/2^4` = 3.0 mg
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