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After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps is
- a)20,000
- b)10,000
- c)7,000
- d)25,000
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
After 300 days, the activity of a radioactive sample is 5000 dps (dis...
Activity decreases 5000 dps to 2500 dps in 150 days
∴ Half life period T1/2 = 150 days
∴ 300 days = 2T1/2
Therefore, initial activity= 5000 x 2T1/2
= 5000 x 2 x 2 = 20000 dps
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Community Answer
After 300 days, the activity of a radioactive sample is 5000 dps (dis...
Given information:
- Activity after 300 days = 5000 dps
- Activity after another 150 days = 2500 dps
To find:
- Initial activity of the sample in dps
Explanation:
Let's assume the initial activity of the radioactive sample after 0 days is A0 (in dps).
Step 1: Calculate the decay constant (λ)
We know that the activity of a radioactive substance decreases exponentially with time according to the equation:
A(t) = A0 * e^(-λt)
Where:
- A(t) is the activity at time t
- A0 is the initial activity
- e is the base of natural logarithm
- λ is the decay constant
To find the decay constant (λ), we can use the given information:
A(300) = A0 * e^(-λ * 300) = 5000
A(450) = A0 * e^(-λ * 450) = 2500
Dividing the second equation by the first equation, we get:
(5000 / 2500) = (A0 * e^(-λ * 450)) / (A0 * e^(-λ * 300))
2 = e^(-λ * 450) / e^(-λ * 300)
Using the property of exponents, we can write:
2 = e^(-λ * 450 - (-λ * 300))
2 = e^(-λ * 150)
Taking the natural logarithm (ln) on both sides, we have:
ln(2) = -λ * 150
Solving for λ, we get:
λ = -ln(2) / 150
Step 2: Calculate the initial activity (A0)
Using the given information, we can substitute the values into the equation:
A(300) = A0 * e^(-λ * 300) = 5000
Substituting the value of λ, we have:
A0 * e^((-ln(2) / 150) * 300) = 5000
Simplifying the equation, we get:
A0 * e^(-2ln(2)) = 5000
Using the property of exponents, we can write:
A0 * (e^ln(2))^(-2) = 5000
A0 * 2^(-2) = 5000
A0 * 1/4 = 5000
A0 = 5000 * 4
Calculating the value, we have:
A0 = 20,000 dps
Therefore, the initial activity of the sample is 20,000 dps. Hence, the correct answer is option 'A'.
- Activity after 300 days = 5000 dps
- Activity after another 150 days = 2500 dps
To find:
- Initial activity of the sample in dps
Explanation:
Let's assume the initial activity of the radioactive sample after 0 days is A0 (in dps).
Step 1: Calculate the decay constant (λ)
We know that the activity of a radioactive substance decreases exponentially with time according to the equation:
A(t) = A0 * e^(-λt)
Where:
- A(t) is the activity at time t
- A0 is the initial activity
- e is the base of natural logarithm
- λ is the decay constant
To find the decay constant (λ), we can use the given information:
A(300) = A0 * e^(-λ * 300) = 5000
A(450) = A0 * e^(-λ * 450) = 2500
Dividing the second equation by the first equation, we get:
(5000 / 2500) = (A0 * e^(-λ * 450)) / (A0 * e^(-λ * 300))
2 = e^(-λ * 450) / e^(-λ * 300)
Using the property of exponents, we can write:
2 = e^(-λ * 450 - (-λ * 300))
2 = e^(-λ * 150)
Taking the natural logarithm (ln) on both sides, we have:
ln(2) = -λ * 150
Solving for λ, we get:
λ = -ln(2) / 150
Step 2: Calculate the initial activity (A0)
Using the given information, we can substitute the values into the equation:
A(300) = A0 * e^(-λ * 300) = 5000
Substituting the value of λ, we have:
A0 * e^((-ln(2) / 150) * 300) = 5000
Simplifying the equation, we get:
A0 * e^(-2ln(2)) = 5000
Using the property of exponents, we can write:
A0 * (e^ln(2))^(-2) = 5000
A0 * 2^(-2) = 5000
A0 * 1/4 = 5000
A0 = 5000 * 4
Calculating the value, we have:
A0 = 20,000 dps
Therefore, the initial activity of the sample is 20,000 dps. Hence, the correct answer is option 'A'.
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After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps isa)20,000b)10,000c)7,000d)25,000Correct answer is option 'A'. Can you explain this answer?
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After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps isa)20,000b)10,000c)7,000d)25,000Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps isa)20,000b)10,000c)7,000d)25,000Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps isa)20,000b)10,000c)7,000d)25,000Correct answer is option 'A'. Can you explain this answer?.
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