Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 min.
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Q: Why do we use 0.67 and 0.33 in the formula instead of 33 and 67?

The radioactive decay law $N(t) = N_0 e^{-\lambda t}$ calculates the amount of substance *remaining* at time $t$, not the amount that has decayed. If $33\%$ has decayed, $67\%$ remains, hence we use $0.67 N_0$.

Q: Is 0.67 divided by 0.33 exactly 2?

Mathematically, $0.67 / 0.33 = 2.0303...$, which is very close to $2$. In JEE physics problems, $33\%$ and $67\%$ are standard approximations for the fractions $1/3$ and $2/3$. Using the exact fractions gives exactly $2$.

Q: Can we solve this problem without using logarithms?

Absolutely! If you recognize that $33\%$ decay leaves $2/3$ of the sample, and $67\%$ decay leaves $1/3$ of the sample, you can see that the amount has halved (from $2/3$ to $1/3$). The time it takes for any radioactive sample to halve is exactly one half-life.

Q: What if the percentages were 20% and 80% instead?

If the percentages were different, the ratio of remaining amounts would not be $2$. For example, $80\%$ remaining and $20\%$ remaining would give a ratio of $4$. You would then use $\ln(4)$ and need the exact value of the decay constant $\lambda$ to find the time.

Q: Why is the decay constant $\lambda$ the same for both states?

The decay constant $\lambda$ is an intrinsic property of the specific radioactive isotope. It does not change over time or depend on how much of the sample is left.

Question

Calculate the time interval between $33\%$ decay and $67\%$ decay if half-life of a substance is $20	ext{ min}$.

Options:

(A) $60	ext{ min}$

(B) $20	ext{ min}$

(C) $40	ext{ min}$

(D) $13	ext{ min}$

Sigma Percentile · Accepted Answer

### The Elegance of Radioactive Decay: Finding the Time Interval

Radioactive decay is one of the most beautifully predictable phenomena in physics. Even though we can never know when a single nucleus will decay, the collective behavior of billions of nuclei follows a strict mathematical law. Let's dive into a classic problem that tests not just your algebra, but your physical intuition.

### Analyzing the Setup

The fundamental law of radioactive decay states that the number of undecayed nuclei $N(t)$ at any time $t$ is given by:

$$N(t) = N_0 e^{-\lambda t}$$

Here, $N_0$ is the initial number of nuclei, and $\lambda$ is the decay constant. A common trap students fall into is plugging the *decayed* percentage directly into this formula. Remember, $N(t)$ represents what is **left**, not what is gone!

When $33\%$ of the substance has decayed, the amount remaining is $100\% - 33\% = 67\%$. Let's call the time it takes to reach this state $t_1$. We can write:

$$0.67 N_0 = N_0 e^{-\lambda t_1}$$

Taking the natural logarithm of both sides gives us:

$$\ln(0.67) = -\lambda t_1$$

Similarly, when $67\%$ of the substance has decayed, the amount remaining is $100\% - 67\% = 33\%$. Let's call this time $t_2$. The equation becomes:

$$0.33 N_0 = N_0 e^{-\lambda t_2}$$

$$\ln(0.33) = -\lambda t_2$$

### The Master Equation

We are looking for the time interval between these two events, which is $\Delta t = t_2 - t_1$. To find this, we can subtract the second logarithmic equation from the first:

$$\lambda t_2 - \lambda t_1 = \ln(0.67) - \ln(0.33)$$

Factoring out $\lambda$ and using the logarithmic property $\ln a - \ln b = \ln(a/b)$, we get:

$$\lambda (t_2 - t_1) = \ln\left(\frac{0.67}{0.33}ight)$$

### Final Calculation and The Intuitive Shortcut

Now, look closely at the fraction $\frac{0.67}{0.33}$. It is approximately equal to $2$. This is not a coincidence; it is a deliberate design of the problem! Substituting this back, we have:

$$\lambda (t_2 - t_1) \approx \ln 2$$

$$t_2 - t_1 = \frac{\ln 2}{\lambda}$$

Does the expression $\frac{\ln 2}{\lambda}$ look familiar? It is the exact definition of the **half-life** ($t_{1/2}$) of a radioactive substance! Since the problem states that the half-life is $20	ext{ min}$, our time interval is simply:

$$\Delta t = 20	ext{ min}$$

**The Intuitive Shortcut:**
Could we have solved this without writing a single equation? Yes! Notice that $33\%$ decay leaves roughly $\frac{2}{3}$ of the sample, and $67\%$ decay leaves roughly $\frac{1}{3}$ of the sample. To go from $\frac{2}{3}$ to $\frac{1}{3}$, the amount of substance must be exactly halved. By definition, the time required for a radioactive sample to halve its quantity is one half-life. Therefore, the time interval is exactly one half-life, which is $20	ext{ min}$. Elegance at its finest!

Animated Solution for Physics - Atoms and Nuclei: Calculate the time interval between $33%$ decay and $67%$ decay if half-life of a substance is $20 min$ .

Select Answer:

Visualized Solution

Radioactive Decay Law

First State: $33%$ Decay

Second State: $67%$ Decay

Finding the Time Interval

Simplifying the Ratio

Relating to Half-Life

The Intuitive Shortcut

The Sigma Insight: Radioactivity

The Elegance of Radioactive Decay

Analyzing the Setup

The Master Equation

Final Calculation and The Intuitive Shortcut

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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; } Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 min.