In given time t=0, Activity of two radioactive substances A and B are equal. After time t, the ratio

Q: Why do we assume $R_{0A} = R_{0B}$?

The question explicitly states that at time $t=0$, the activities of the two radioactive substances $A$ and $B$ are equal. Therefore, their initial activities are the same.

Q: What is the relationship between half-life and decay constant?

The half-life $T_{1/2}$ and the decay constant $\lambda$ are inversely proportional, related by the equation $T_{1/2} = \frac{\ln 2}{\lambda}$.

Q: Why does the ratio decrease exponentially?

Because both activities decay exponentially over time. When you divide two exponential functions with the same base, the result is another exponential function whose exponent is the difference of the original exponents.

Q: Can we solve this without finding $\lambda_A$ first?

You can keep $\lambda_A$ as a variable in your equations, but you will eventually need to substitute its value (derived from the given half-life of A) to find the numerical value of $\lambda_B$.

Q: What does a larger decay constant mean physically?

A larger decay constant means the radioactive substance decays at a faster rate. Consequently, it will have a shorter half-life, meaning it takes less time for half of the sample to disintegrate.

Question

In given time $t=0$, Activity of two radioactive substances $A$ and $B$ are equal. After time $t$, the ratio of their activities $\frac{R_B}{R_A}$ decreases according to $e^{-3t}$. If the half life of $A$ is $\ln 2$, the half-life of $B$ will be

Options:

(A) $4\ln 2$

(B) $\frac{\ln 2}{4}$

(C) $\frac{\ln 2}{2}$

(D) $2\ln 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Let's start by looking at the initial conditions. At time $t = 0$, the activities of both radioactive substances, $A$ and $B$, are perfectly equal. We can call this initial activity $R_0$.

Now, according to the radioactive decay law, the activity at any time $t$ follows an exponential curve. It is given by the master equation:

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

where $\lambda$ is the decay constant. This equation tells us exactly how fast a radioactive sample is disintegrating at any given moment.

### Finding the Decay Constant of A
We are given a crucial piece of information: the half-life of substance $A$ is $\ln 2$. We know the standard formula relating half-life and the decay constant is:

$$(T_{1/2})_A = \frac{\ln 2}{\lambda_A}$$

Equating this to the given value, we get:

$$\frac{\ln 2}{\lambda_A} = \ln 2$$

Solving this simple equation, we easily find that the decay constant for $A$, $\lambda_A$, is exactly $1$.

### The Ratio of Activities
Next, let's find the ratio of their activities at any time $t$. By dividing $R_B$ by $R_A$, the initial activity $R_0$ beautifully cancels out:

$$\frac{R_B(t)}{R_A(t)} = \frac{R_0 e^{-\lambda_B t}}{R_0 e^{-\lambda_A t}}$$

Using the fundamental properties of exponents, we can combine the terms in the numerator and denominator:

$$\frac{R_B(t)}{R_A(t)} = e^{-(\lambda_B - \lambda_A)t}$$

### Comparing Exponents and Final Calculation
The problem states this ratio decreases according to $e^{-3t}$. By comparing the exponents of our derived ratio and the given ratio, we can clearly see that:

$$\lambda_B - \lambda_A = 3$$

Since we already calculated $\lambda_A = 1$, let's substitute it here. This gives us:

$$\lambda_B = 1 + 3 = 4$$

So, substance $B$ decays four times faster than $A$! Finally, we need to find the half-life of substance $B$. Using the half-life formula again, $(T_{1/2})_B$ is $\frac{\ln 2}{\lambda_B}$. Substituting $\lambda_B = 4$, we get our final answer:

**$$(T_{1/2})_B = \frac{\ln 2}{4}$$**

Select Answer:

Visualized Solution

Initial Activity

Activity Law

Decay Constant of A

Ratio of Activities

Comparing Exponents

Decay Constant of B

Half-life of B

The Sigma Insight: Radioactivity

Analyzing the Setup Let's start by looking at the initial conditions

Finding the Decay Constant of A We are given a crucial piece of information: the half-life of substance $A$ is $ln 2$

The Ratio of Activities Next, let's find the ratio of their activities at any time $t$

Comparing Exponents and Final Calculation The problem states this ratio decreases according to $e^{- 3 t}$

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