Problem 67
Question
A radioactive substance having a half-life of 3 days was received in 12 days. It was found that there was only \(3 \mathrm{~g}\) of the isotope in the container. The initial weight of the isotope when packed was (a) \(12 \mathrm{~g}\) (b) \(24 \mathrm{~g}\) (c) \(48 \mathrm{~g}\) (d) \(96 \mathrm{~g}\)
Step-by-Step Solution
Verified Answer
The initial weight of the isotope was 48 grams (option c).
1Step 1: Determine the Decay Formula for Half-Life
The decay of a radioactive substance is given by the formula \( N(t) = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{t}{T}} \), where \( N(t) \) is the amount remaining, \( N_0 \) is the initial amount, \( t \) is the time elapsed, and \( T \) is the half-life of the substance.
2Step 2: Identify Known Values
From the exercise, the half-life \( T = 3 \) days, time elapsed \( t = 12 \) days, and the remaining amount \( N(t) = 3 \) grams. We need to find \( N_0 \), the initial amount.
3Step 3: Set Up the Equation
Using the decay formula, substitute the known values: \( 3 = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{12}{3}} \). Simplifying the exponent gives \( 3 = N_0 \cdot \left( \frac{1}{2} \right)^4 \).
4Step 4: Solve for Initial Amount \( N_0 \)
The expression \( \left( \frac{1}{2} \right)^4 \) equals \( \frac{1}{16} \). Thus, \( 3 = N_0 \cdot \frac{1}{16} \). To isolate \( N_0 \), multiply both sides by 16: \( N_0 = 3 \times 16 \).
5Step 5: Calculate \( N_0 \) and Choose the Correct Answer
Calculate \( N_0 = 48 \) grams. Comparing with the options, the initial weight of the isotope was 48 grams, option (c).
Key Concepts
Half-Life CalculationExponential Decay FormulaInitial Quantity of Isotope
Half-Life Calculation
The concept of half-life is a fundamental principle in radioactive decay measurements. It describes the time required for a quantity of a radioactive substance to reduce to half its initial amount. In the context of our exercise, the half-life of the substance is given as 3 days. This means that every 3 days, the remaining amount of the isotope will be halved.
Understanding half-life helps predict how long it will take for a radioactive material to decay to a safer, less radioactive level.
Understanding half-life helps predict how long it will take for a radioactive material to decay to a safer, less radioactive level.
- If a radioactive substance starts with a certain number of atoms, in one half-life, that number will be reduced by 50%.
- This does not mean the substance is half gone, just that half of its radioactive activity has decayed.
Exponential Decay Formula
Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In radioactive decay, this is captured mathematically by the exponential decay formula: \[N(t) = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{t}{T}}\]where:
Understanding the exponential decay formula equips you to calculate the amount of substance remaining at any given time.
- \(N(t)\) is the remaining quantity of the substance after time \(t\).
- \(N_0\) is the initial quantity.
- \(T\) is the half-life of the substance.
Understanding the exponential decay formula equips you to calculate the amount of substance remaining at any given time.
Initial Quantity of Isotope
Determining the initial quantity of an isotope is critical in many scientific applications, from radiometric dating to medical diagnostics. The initial quantity, often denoted as \(N_0\), refers to the amount of the substance before any decay has occurred.
This value can be found using the exponential decay formula by rearranging it to solve for \(N_0\) if the remaining amount and elapsed time are known.
This value can be found using the exponential decay formula by rearranging it to solve for \(N_0\) if the remaining amount and elapsed time are known.
- In the problem, we find that after 12 days, only 3 grams remain.
- Using the formula \(3 = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{12}{3}}\), we determine \(N_0\).
- Upon solving, we find that the initial quantity was 48 grams.
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