Two radioactive materials A and B have decay constants 25λ and 16λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be ‘e’ after a time \(\frac{1}{(aλ)}\). The value of a is _____.
Correct Answer: 9
Approach Solution - 1
To solve the problem, we first need to find the time at which the ratio of the number of nuclei of B to A is \(e\). Begin with the decay formula for any radioactive material: \(N(t) = N_0 e^{-\lambda t}\), where \(N_0\) is the initial number of nuclei and \(\lambda\) is the decay constant.
Since both materials initially have the same number of nuclei, \(N_{0A} = N_{0B}\), the number of nuclei at a given time \(t\) will be:
For material A: \(N_A(t) = N_0 e^{-25\lambda t}\)
For material B: \(N_B(t) = N_0 e^{-16\lambda t}\)
The given condition is that the ratio \( \frac{N_B(t)}{N_A(t)} = e \). Substitute the expressions:
\(\frac{N_0 e^{-16\lambda t}}{N_0 e^{-25\lambda t}} = e\)
This simplifies to:
\(e^{9\lambda t} = e\)
Taking the natural logarithm on both sides gives:
\(9\lambda t = 1\)
Thus, \(t = \frac{1}{9\lambda}\)
We need to identify the value of \(a\) in \(\frac{1}{a\lambda}\) such that the solution fits within the expected range.
Comparing \(t = \frac{1}{a\lambda}\) with \(t = \frac{1}{9\lambda}\), we find \(a = 9\).
Confirming the range: The value of \(a\) we calculated is \(9\), which coincides with the given range of 9 to 9.
Therefore, the value of \(a\) is 9.
Approach Solution -2
NA=N0e−25λt
NB=N0e−16λt
\(\frac{N_B}{N_A}\)=e9λt
\(t=\frac{1}{9λ}\)
Given that, ratio of the number of nuclei of B to that of A will be e after a time \(\frac{1}{9λ}\)
On comparing, a = 9
So, the answer is 9.
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