Problem 46
Question
There are two radioactive substances \(A\) and \(B\). Decay constant of \(B\) is two times that of \(A\). Initially, both have equal number of nuclei. After \(n\) half-lives of \(A\), rate of disintegration of both are equal. The value of \(n\) is (a) 4 (b) 2 (c) 1 (d) 5
Step-by-Step Solution
Verified Answer
n = 1
1Step 1: Understand the Decay Law
The number of remaining nuclei in a radioactive substance after time \( t \) is given by the formula \( N(t) = N_0 \cdot e^{-\lambda t} \), where \( \lambda \) is the decay constant and \( N_0 \) is the initial number of nuclei.
2Step 2: Identify Decay Constants
Let the decay constant of substance \( A \) be \( \lambda_A \). Since the decay constant of \( B \) is two times that of \( A \), we have \( \lambda_B = 2\lambda_A \).
3Step 3: Calculate Rate of Disintegration
The rate of disintegration or activity for a substance at time \( t \) is given by \( R(t) = \lambda \cdot N(t) \). For \( A \) and \( B \), this becomes \( R_A(t) = \lambda_A \cdot N_A(t) \) and \( R_B(t) = \lambda_B \cdot N_B(t) \), respectively.
4Step 4: Apply Initial Conditions
Initially, both substances have the same number of nuclei, \( N_0 \). Thus, \( N_A(0) = N_B(0) = N_0 \).
5Step 5: Consider After n Half-Lives of A
The time after \( n \) half-lives of \( A \) is \( t = n \cdot \frac{\ln(2)}{\lambda_A} \). The number of remaining nuclei in \( A \) after \( n \) half-lives is \( N_A(t) = N_0 \cdot e^{-\lambda_A t} \). For \( B \), it's \( N_B(t) = N_0 \cdot e^{-2\lambda_A t} \).
6Step 6: Set Equations for Equal Disintegration Rates
According to the problem, the rates of disintegration of \( A \) and \( B \) are equal after \( n \) half-lives of \( A \). Therefore, \( \lambda_A \cdot N_A(t) = \lambda_B \cdot N_B(t) \). Substitute \( \lambda_B = 2\lambda_A \) into the equation: \( \lambda_A \cdot N_0 \cdot e^{-\lambda_A t} = 2\lambda_A \cdot N_0 \cdot e^{-2\lambda_A t} \).
7Step 7: Simplify and Solve for n
Cancel \( \lambda_A \cdot N_0 \) from both sides, resulting in \( e^{-\lambda_A t} = 2 \cdot e^{-2\lambda_A t} \). Rearranging gives \( e^{\lambda_A t} = 2 \), and taking the logarithm gives \( \lambda_A t = \ln(2) \). Substitute \( t = n \cdot \frac{\ln(2)}{\lambda_A} \) into the equation to solve for \( n \): \( n \cdot \ln(2) = \ln(2) \) which simplifies to \( n = 1 \).
Key Concepts
Decay ConstantHalf-LifeRate of Disintegration
Decay Constant
The decay constant \( \lambda \) is a fundamental concept that defines the rate at which a radioactive substance undergoes decay. It is a measure of the probability per unit time that a single nucleus will decay. The larger the decay constant, the quicker the substance will decay.
- It is expressed in units of inverse time, such as \( \text{s}^{-1} \) or \( \text{year}^{-1}\).
- In mathematical terms, it is used in the equation \( N(t) = N_0 \cdot e^{-\lambda t} \), where \( N(t) \) is the number of remaining nuclei at time \( t \).
Half-Life
The half-life of a radioactive substance is the time required for half of its nuclei to decay. It serves as an intuitive way to understand how quickly a substance undergoes radioactive decay.
- The formula for half-life \( T_{1/2} \) is \( T_{1/2} = \frac{\ln(2)}{\lambda} \).
- Each half-life reduces the number of undecayed nuclei by half, a situation that occurs repeatedly over the lifetime of a substance.
Rate of Disintegration
The rate of disintegration, also known as the activity, is the number of decays per unit time in a sample of radioactive material. It is directly related to the decay constant and the number of undecayed nuclei.
- The formula is \( R(t) = \lambda \cdot N(t) \).
- This means the activity changes as the number of radioactive nuclei decreases over time.
Other exercises in this chapter
Problem 45
The electron in a hydrogen atom makes a transition \(n_{1} \rightarrow n_{2}\) where \(n_{1}\) and \(n_{2}\) are the principal quantum numbers of the two states
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The ratio of the energies of the hydrogen atom in its first to second excited states is (a) \(9 / 4\) (b) \(4 / 1\) (c) \(8 / 1\) (d) \(1 / 8\)
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A hydrogen atom initially in the ground level absorbs a photon, which excites it to the \(n=4\) level. Determine the wavelength and frequency of photon. (a) \(9
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The binding energy of an electron in the ground state of He is equal to \(24.6 \mathrm{eV}\). The energy required to remove both the electrons is (a) \(49.2 \ma
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