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A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.
Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)
that will decay in the first 80 s after preparation of the sample is
Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)
that will decay in the first 80 s after preparation of the sample is
Correct answer is '4'. Can you explain this answer?
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A freshly prepared sample of a radioisotope of half-life 1386 s has ac...
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Given information:
- Half-life of the radioisotope = 1386 s
- Activity of the sample = 103 disintegrations per second
- ln 2 = 0.693
To Find:
The fraction of the initial number of nuclei that will decay in the first 80 s after preparation of the sample.
Solution:
Step 1: Find the decay constant (λ)
The decay constant (λ) can be calculated using the formula:
λ = ln(2) / half-life
Plugging in the values, we get:
λ = 0.693 / 1386
λ ≈ 0.0004996 s⁻¹
Step 2: Find the number of remaining nuclei after 80 s
The number of remaining nuclei (N) can be calculated using the formula:
N = N₀ * e^(-λt)
Where:
N₀ = initial number of nuclei
t = time
Plugging in the values, we get:
N = N₀ * e^(-0.0004996 * 80)
Step 3: Find the fraction of decay
The fraction of decay can be calculated using the formula:
Fraction of decay = (N₀ - N) / N₀ * 100
Plugging in the values, we get:
Fraction of decay = (N₀ - N) / N₀ * 100
Fraction of decay = (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Step 4: Calculate the fraction of decay
Substitute the calculated values of N₀ and N into the formula:
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Final Answer:
The fraction of the initial number of nuclei that will decay in the first 80 s after preparation of the sample is approximately 4%.
- Half-life of the radioisotope = 1386 s
- Activity of the sample = 103 disintegrations per second
- ln 2 = 0.693
To Find:
The fraction of the initial number of nuclei that will decay in the first 80 s after preparation of the sample.
Solution:
Step 1: Find the decay constant (λ)
The decay constant (λ) can be calculated using the formula:
λ = ln(2) / half-life
Plugging in the values, we get:
λ = 0.693 / 1386
λ ≈ 0.0004996 s⁻¹
Step 2: Find the number of remaining nuclei after 80 s
The number of remaining nuclei (N) can be calculated using the formula:
N = N₀ * e^(-λt)
Where:
N₀ = initial number of nuclei
t = time
Plugging in the values, we get:
N = N₀ * e^(-0.0004996 * 80)
Step 3: Find the fraction of decay
The fraction of decay can be calculated using the formula:
Fraction of decay = (N₀ - N) / N₀ * 100
Plugging in the values, we get:
Fraction of decay = (N₀ - N) / N₀ * 100
Fraction of decay = (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Step 4: Calculate the fraction of decay
Substitute the calculated values of N₀ and N into the formula:
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Fraction of decay ≈ (N₀ - N) / N₀ * 100
Final Answer:
The fraction of the initial number of nuclei that will decay in the first 80 s after preparation of the sample is approximately 4%.
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A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer?
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A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer?.
A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second.Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage)that will decay in the first 80 s after preparation of the sample isCorrect answer is '4'. Can you explain this answer?.
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