Question

In a radioactive material the activity at time t₁ is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is \(\lambda\), then :

• A. R₁ = R₂ \(\text{e}^{-\lambda(t_1-t_2)}\)
• B. R₁= R₂ \(\text{e}^{\lambda(t_1-t_2)}\)
• C. R₁= R₂ \(\text{e}^{\bigg(\frac{t_2}{t_1}\bigg)}\)
• D. R₁ = R₂

Answer 1

The Correct Option is A

 In this problem, we are dealing with the radioactive decay activity at two different times, \(t_1\) and \(t_2\), with corresponding activities \(R_1\) and \(R_2\) respectively. We are required to find the relationship between these activities given the decay constant \(\lambda\).

Radioactive decay follows an exponential decay law. The activity \(R\) at any time \(t\) is given by:

\(R = R_0 \text{e}^{-\lambda t}\)

where:

• \(R_0\) is the initial activity at time \(t = 0\).
• \(\lambda\) is the decay constant.
• \(t\) is the time elapsed.

 

Given that the activity at time \(t_1\) is \(R_1\) and at a later time \(t_2\) it is \(R_2\), we can write:

\(R_1 = R_0 \text{e}^{-\lambda t_1}\)

\(R_2 = R_0 \text{e}^{-\lambda t_2}\)

To find the relationship between \(R_1\) and \(R_2\), divide the first equation by the second:

\(\frac{R_1}{R_2} = \frac{R_0 \text{e}^{-\lambda t_1}}{R_0 \text{e}^{-\lambda t_2}}\)

Simplifying the above equation, we have:

\(\frac{R_1}{R_2} = \text{e}^{-\lambda t_1 + \lambda t_2}\)

\(\frac{R_1}{R_2} = \text{e}^{\lambda (t_2 - t_1)}\)

Therefore, rearranging gives:

\(R_1 = R_2 \times \text{e}^{\lambda (t_2 - t_1)}\)

To match the given format of the options, we can also write this as:

\(R_1 = R_2 \times \text{e}^{-\lambda (t_1 - t_2)}\)

Thus, option \(R_1 = R_2 \text{e}^{-\lambda(t_1-t_2)}\) is correct. This option correctly reflects the decrease in activity due to decay over the time period.