Question

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 _____.

Answer 1

Correct Answer: 9

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.

Answer 2

N_A=N₀e^−25λt
N_B=N₀e^−16λt
\(\frac{N_B}{N_A}\)=e^9λt
\(t=\frac{1}{9λ}\)
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.