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Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is
- a)1 /3
- b)1 /2
- c)1/8
- d)1 /4
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Plutonium decays with a half-life of 24000 years. If the plutonium is ...
The amount of plotinium after a time period of 72000 if the half life is 24000 will be
the initial amount x would be reduced to x/2 , in 24000 yrs
then it would lessen to x/4 in the next 24000yrs
and then to x/8 in the next 24000 yrs
that is it will reduce to x/8 in the next 72000yrs starting from x .
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Plutonium decays with a half-life of 24000 years. If the plutonium is ...
Solution:
The formula for the decay of a radioactive substance is given by:
N(t) = N₀e^(-λt)
where,
N(t) = Number of radioactive nuclei at time t
N₀ = Initial number of radioactive nuclei
λ = Decay constant
t = Time
Given,
Half-life of Plutonium = 24000 years
So, λ = ln(2) / T₁/₂ = ln(2) / 24000
After 72000 years, the fraction of Plutonium that remains can be calculated as follows:
N(72000) = N₀e^(-λ*72000)
Fraction of Plutonium that remains = N(72000) / N₀
Now, we need to simplify the expression for N(72000) / N₀:
N(72000) / N₀ = e^(-λ*72000)
= e^(-ln(2)*72000/24000)
= e^(-3ln(2))
= e^ln(2^-3)
= 2^-3
= 1/8
Therefore, the fraction of Plutonium that remains after 72000 years is 1/8, which is option (c).
The formula for the decay of a radioactive substance is given by:
N(t) = N₀e^(-λt)
where,
N(t) = Number of radioactive nuclei at time t
N₀ = Initial number of radioactive nuclei
λ = Decay constant
t = Time
Given,
Half-life of Plutonium = 24000 years
So, λ = ln(2) / T₁/₂ = ln(2) / 24000
After 72000 years, the fraction of Plutonium that remains can be calculated as follows:
N(72000) = N₀e^(-λ*72000)
Fraction of Plutonium that remains = N(72000) / N₀
Now, we need to simplify the expression for N(72000) / N₀:
N(72000) / N₀ = e^(-λ*72000)
= e^(-ln(2)*72000/24000)
= e^(-3ln(2))
= e^ln(2^-3)
= 2^-3
= 1/8
Therefore, the fraction of Plutonium that remains after 72000 years is 1/8, which is option (c).
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Plutonium decays with a half-life of 24000 years. If the plutonium is ...
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Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is a)1 /3b)1 /2c)1/8d)1 /4Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is a)1 /3b)1 /2c)1/8d)1 /4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is a)1 /3b)1 /2c)1/8d)1 /4Correct answer is option 'C'. Can you explain this answer?.
Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is a)1 /3b)1 /2c)1/8d)1 /4Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is a)1 /3b)1 /2c)1/8d)1 /4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Plutonium decays with a half-life of 24000 years. If the plutonium is stored for 72000 years, then the fraction of plutonium that remains is a)1 /3b)1 /2c)1/8d)1 /4Correct answer is option 'C'. Can you explain this answer?.
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