Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two h

Q: Why is the 4 days half-life not used in the calculation?

The question asks for the probability after 'two half-lives'. Since the time interval is already given in terms of half-lives, the absolute value of the half-life (4 days) is redundant information.

Q: Why isn't the probability of decay in two half-lives just $1/2 + 1/2 = 1$?

Decay is a continuous, exponential process, not a simple linear addition. In the second half-life, only half of the *remaining* nuclei decay, not half of the original amount.

Q: Does this probability apply to a single nucleus?

Yes! The macroscopic fraction of a large sample that decays is mathematically identical to the quantum mechanical probability that any single individual nucleus will decay in that same time frame.

Q: What is the probability of a nucleus surviving after two half-lives?

The survival probability is the fraction remaining, which is $(1/2)^2 = 1/4$.

Q: How do we calculate the probability for a time that is not an exact multiple of the half-life?

You use the exponential decay law: $P(\text{decay}) = 1 - e^{-\lambda t}$, where $\lambda = \ln(2)/T_{1/2}$.

Question

Half-life of a radioactive substance $A$ is $4$ days. The probability that a nucleus will decay in two half-lives is

Options:

(A) $\frac{1}{4}$

(B) $\frac{3}{4}$

(C) $\frac{1}{2}$

(D) $1$

Sigma Percentile · Accepted Answer

### The Illusion of Extra Information

Welcome to a classic trap set by examiners! The question begins by stating that the half-life of a radioactive substance is $4$ days. But then, it asks for the probability of decay in exactly *two half-lives*. Do we actually need the $4$ days to solve this? Absolutely not!

The time interval is already provided in terms of the half-life itself. Whether the half-life is $4$ days, $4$ years, or $4$ milliseconds, the fraction of the sample that decays after two half-lives remains mathematically identical. The $4$ days is merely a distractor designed to make you overthink.

### Decoding the Half-Life

Let's break down the journey of our radioactive sample. Imagine a block of $N_0$ active nuclei. By definition, a half-life ($T_{1/2}$) is the time required for exactly half of the radioactive nuclei in a given sample to undergo decay.

After the first half-life passes, the math is straightforward:

$$N_1 = \frac{N_0}{2}$$

We are left with exactly half of our original sample. The other half has decayed into a daughter element.

### The Journey of Two Half-Lives

Now we move into the second half-life, and this is where students often make a critical conceptual error. They assume that since half decayed in the first half-life, the other half will decay in the second, leaving nothing.

**This is incorrect.** Radioactive decay is an exponential process, not a linear one. The decay always happens on the *remaining* nuclei, not the original amount. So, during the second half-life, half of the *remaining* $N_0/2$ nuclei will decay.

$$N_2 = \frac{1}{2} 	imes \frac{N_0}{2} = \frac{N_0}{4}$$

After two half-lives, we are left with one-fourth of the initial sample.

### The Quantum Connection: From Sample to Single Nucleus

To find the number of nuclei that have decayed, we simply subtract the remaining nuclei from the initial amount:

$$	ext{Decayed Nuclei} = N_0 - \frac{N_0}{4} = \frac{3N_0}{4}$$

But wait, the question asks for the probability of a *single* nucleus decaying. Here is the beautiful bridge between the macroscopic and the quantum world: The macroscopic fraction of a large sample that decays is mathematically identical to the quantum mechanical probability that any single individual nucleus will decay in that same time frame.

### The Final Calculation

The probability of decay is the ratio of decayed nuclei to the initial number of nuclei:

$$P(	ext{decay}) = \frac{\frac{3N_0}{4}}{N_0} = \frac{3}{4}$$

Therefore, the probability that any given nucleus decays within two half-lives is $\frac{3}{4}$.

### The General Pattern

What if the question asked for the probability of decay after $n$ half-lives? You can generalize this logic beautifully. The fraction remaining after $n$ half-lives is always $(1/2)^n$. Therefore, the probability of decay is simply:

$$P(	ext{decay}) = 1 - \left(\frac{1}{2}ight)^n$$

Always look for the general pattern, and you will never be tricked by extra numbers again!

Animated Solution for Physics - Atoms and Nuclei: Half-life of a radioactive substance $A$ is $4$ days. The probability that a nucleus will decay in two half-lives is

Select Answer:

Visualized Solution

$Initial State$

$First Half-Life$

$Second Half-Life$

$Probability of Decay$

$Calculating Decayed Fraction$

$Final Answer$

$The Way Forward$

The Sigma Insight: Radioactivity

The Illusion of Extra Information

Decoding the Half-Life

The Journey of Two Half-Lives

The Quantum Connection

The Final Calculation

The General Pattern

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Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is