A sample of radioactive material A, that has an activity of 10 mCi (1 Ci = 3.7 x 10^10 decays/s) has

Q: Why didn't we use the conversion factor $1\text{ Ci} = 3.7 \times 10^{10}\text{ decays/s}$?

Because we are calculating a ratio ($\frac{R_A}{R_B}$). As long as both activities are in the same units (mCi), the units will cancel out completely. Converting them to decays/s would just add unnecessary large numbers to the calculation without changing the final ratio.

Q: Why is the half-life inversely proportional to the decay constant?

The radioactive decay law is $N = N_0 e^{-\lambda t}$. By definition, at one half-life ($t = T_{1/2}$), the number of nuclei $N$ becomes $\frac{N_0}{2}$. Substituting this gives $\frac{1}{2} = e^{-\lambda T_{1/2}}$. Taking the natural log of both sides yields $\ln(2) = \lambda T_{1/2}$, which rearranges to $T_{1/2} = \frac{\ln 2}{\lambda}$.

Q: Could we have found the exact half-lives without looking at the options?

No. The information provided only allows us to find the *ratio* of the half-lives ($\frac{T_A}{T_B} = 4$). There are infinitely many pairs of numbers that have a ratio of 4 (e.g., 4 and 1, 8 and 2, 40 and 10). The options are necessary to pinpoint the specific pair intended by the examiner.

Q: What exactly does 'activity' mean in physical terms?

Activity is the rate at which a radioactive sample decays. It tells you how many individual nuclei are disintegrating every single second. A higher activity means the sample is highly unstable and decaying very rapidly.

Q: I got confused when substituting $N_A = 2N_B$. Why did $N_B$ disappear?

When you set up the ratio $\frac{\lambda_A N_A}{\lambda_B N_B}$ and substitute $2N_B$ for $N_A$, you get $\frac{\lambda_A (2N_B)}{\lambda_B N_B}$. Since $N_B$ is a non-zero quantity present in both the numerator and the denominator, it acts as a common factor and cancels out, leaving just $\frac{2\lambda_A}{\lambda_B}$.

Question

A sample of radioactive material $A$, that has an activity of $10	ext{ mCi}$ ($1	ext{ Ci} = 3.7 	imes 10^{10}	ext{ decays/s}$) has twice the number of nuclei as another sample of a different radioactive material $B$ which has an activity of $20	ext{ mCi}$. The correct choices for half-lives of $A$ and $B$ would, then be respectively

Options:

(A) 20 days and 10 days

(B) 5 days and 10 days

(C) 10 days and 40 days

(D) 20 days and 5 days

Sigma Percentile · Accepted Answer

### Unraveling Radioactive Decay: A Tale of Two Samples

Radioactivity can sometimes feel like a game of invisible numbers, but at its core, it is governed by elegant and straightforward mathematical laws. In this problem, we are presented with two distinct radioactive samples, $A$ and $B$, and we need to deduce their half-lives based on their activities and the number of nuclei they contain. Let's break down the physics and the math step-by-step.

### The Master Equation: Activity

The first concept we need to anchor ourselves to is **Activity**. Activity, denoted by $R$ (or sometimes $A$), is simply the rate at which a radioactive sample decays. Think of it as the 'speedometer' of the radioactive sample. The fundamental law of radioactive decay tells us that this activity is directly proportional to the number of undecayed nuclei $N$ present in the sample.

The equation is:
$$R = \lambda N$$

Here, $\lambda$ is the **decay constant**, a unique fingerprint for every radioactive isotope that tells us the probability of decay per unit time.

### Setting Up the Ratios

We are given the activities for both samples:
- For Sample A: $R_A = 10	ext{ mCi}$
- For Sample B: $R_B = 20	ext{ mCi}$

*(Note: The conversion factor $1	ext{ Ci} = 3.7 	imes 10^{10}	ext{ decays/s}$ is provided, but since we will be dealing with ratios, the units of milliCuries will neatly cancel out, saving us from tedious calculations!)*

The problem also hands us a golden key: Sample $A$ has twice the number of nuclei as Sample $B$. Mathematically, this translates to:
$$N_A = 2N_B$$

Now, let's use our master equation to set up a ratio between the two samples. Ratios are incredibly powerful in physics because they allow us to eliminate unknown variables.

$$\frac{R_A}{R_B} = \frac{\lambda_A N_A}{\lambda_B N_B}$$

### The Algebraic Execution

Let's substitute the values we know into our ratio equation:

$$\frac{10}{20} = \frac{\lambda_A (2N_B)}{\lambda_B N_B}$$

Notice the beauty of algebra here: the unknown $N_B$ appears in both the numerator and the denominator, so it cancels out completely!

$$\frac{1}{2} = \frac{2\lambda_A}{\lambda_B}$$

By rearranging the terms, we can isolate the ratio of the decay constants:

$$\frac{\lambda_A}{\lambda_B} = \frac{1}{4}$$

### Connecting Decay Constant to Half-Life

We have the ratio of the decay constants, but the question asks for the half-lives. How are they connected? The half-life $T_{1/2}$ is the time required for half of the radioactive nuclei in a sample to decay. It is inversely proportional to the decay constant:

$$T_{1/2} = \frac{\ln 2}{\lambda}$$

Because of this inverse relationship, the ratio of the half-lives will be the exact reciprocal of the ratio of the decay constants:

$$\frac{(T_{1/2})_A}{(T_{1/2})_B} = \frac{\lambda_B}{\lambda_A} = 4$$

### The Final Deduction

We have deduced that the half-life of Sample $A$ must be exactly $4$ times the half-life of Sample $B$. Now, we simply play detective with the given options:

- (a) $20	ext{ days}$ and $10	ext{ days} ightarrow 	ext{Ratio} = 2$
- (b) $5	ext{ days}$ and $10	ext{ days} ightarrow 	ext{Ratio} = 0.5$
- (c) $10	ext{ days}$ and $40	ext{ days} ightarrow 	ext{Ratio} = 0.25$
- (d) $20	ext{ days}$ and $5	ext{ days} ightarrow 	ext{Ratio} = 4$

Only option (d) satisfies our rigorously derived ratio. Thus, the half-lives of $A$ and $B$ are $20	ext{ days}$ and $5	ext{ days}$, respectively.

Select Answer:

Visualized Solution

Visualizing the Samples

The Activity Formula

Relating the Nuclei

Setting up the Ratio

Substituting Values

Solving for Decay Constants

Relating to Half-Life

Finding the Correct Option

The Sigma Insight: Radioactivity

Unraveling Radioactive Decay

The Master Equation

Setting Up the Ratios

The Algebraic Execution

Connecting Decay Constant to Half-Life

The Final Deduction

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