The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20th of the rado

Q: Why did we use $\log_{10}$ instead of natural log?

The ratio of two logarithms of the same base is independent of the base. $\frac{\ln 20}{\ln 2} = \frac{\log_{10} 20}{\log_{10} 2}$. Base 10 logs are easier to remember and calculate.

Q: What is the significance of the given $\log_{10} e = 0.4343$?

It is a conversion factor between natural log and base 10 log ($\ln x = \frac{\log_{10} x}{\log_{10} e}$). However, by taking the ratio of two logs, this factor cancels out, making it unnecessary for the final calculation.

Q: Why is the answer 16.5 days when the exact calculation gives 16.42 days?

In older JEE papers, options were sometimes slightly rounded or based on approximate log values (like $\ln 10 \approx 2.3$). 16.5 is the closest and uniquely correct option.

Q: Can we solve this without knowing log values?

Yes! By estimating powers of 2. $\frac{1}{16}$ remains after 4 half-lives (15.2 days) and $\frac{1}{32}$ after 5 half-lives (19 days). Since $\frac{1}{20}$ is between them, the time must be between 15.2 and 19 days.

Q: Does the initial amount of radon matter?

No, the time taken to reach a specific fraction of the initial amount depends only on the half-life, not on the initial quantity $N_0$, as it cancels out in the equation.

Question

The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20th of the radon sample will remain undecayed, is (given $\log_{10} e = 0.4343$)

Options:

(A) 3.8 days

(B) 16.5 days

(C) 33 days

(D) 76 days

Sigma Percentile · Accepted Answer

### The Exponential Nature of Decay

Imagine holding a sample of radioactive radon. It doesn't just sit there; it is actively transforming, emitting radiation as its unstable nuclei decay into more stable forms.

This process is entirely random for a single nucleus, but for a massive collection of them, it follows a beautifully predictable mathematical law: the law of exponential decay.

The number of undecayed nuclei $N$ at any time $t$ is given by the master equation:
$$N = N_0 e^{-\lambda t}$$

Here, $N_0$ is the initial number of nuclei, and $\lambda$ is the decay constant, which dictates how aggressively the substance decays.

### Decoding the Half-Life

We are given that the half-life $t_{1/2}$ of radon is $3.8$ days.

The half-life is the exact time required for half of the radioactive nuclei in a sample to undergo decay.

Mathematically, the decay constant $\lambda$ is intimately tied to the half-life through the relation:
$$\lambda = \frac{\ln 2}{t_{1/2}}$$

This means we already have all the information we need to find the decay constant, and consequently, the behavior of the entire sample over time.

### Setting Up the Target

Our mission is to find the time $t$ when exactly $\frac{1}{20}$th of the original radon sample remains undecayed.

We can translate this physical condition into our mathematical framework by setting $N = \frac{N_0}{20}$.

Substituting this into our decay equation, we get:
$$\frac{N_0}{20} = N_0 e^{-\lambda t}$$

The $N_0$ terms gracefully cancel out, leaving us with a pure exponential equation:
$$\frac{1}{20} = e^{-\lambda t}$$

### The Power of Logarithms

To extract the time $t$ from the exponent, we must apply the natural logarithm to both sides of the equation.

$$\ln\left(\frac{1}{20}ight) = -\lambda t$$

Using the properties of logarithms, $\ln(1/20)$ is simply $-\ln(20)$. The negative signs on both sides cancel out perfectly:
$$\ln(20) = \lambda t$$

Now, we substitute our earlier expression for the decay constant $\lambda$:
$$\ln(20) = \left(\frac{\ln 2}{3.8}ight) t$$

### The Final Calculation

Isolating $t$, we arrive at the final expression:
$$t = 3.8 	imes \frac{\ln 20}{\ln 2}$$

Here is a brilliant mathematical trick: the ratio of two natural logarithms is identical to the ratio of their base-10 counterparts!
$$t = 3.8 	imes \frac{\log_{10} 20}{\log_{10} 2}$$

We know that $\log_{10} 20 = \log_{10}(2 	imes 10) = \log_{10} 2 + 1 \approx 1.3010$.
And $\log_{10} 2 \approx 0.3010$.

Plugging these standard values in:
$$t = 3.8 	imes \frac{1.3010}{0.3010} \approx 16.42 	ext{ days}$$

The closest option provided is **16.5 days**, making it our definitive answer.

### The Pro-Tip: Smart Estimation

In competitive exams like JEE, time is your most valuable asset. You could have solved this without calculating a single logarithm!

Think in terms of half-lives. After 4 half-lives, the remaining fraction is $\frac{1}{2^4} = \frac{1}{16}$. This takes $4 	imes 3.8 = 15.2$ days.

After 5 half-lives, the remaining fraction is $\frac{1}{2^5} = \frac{1}{32}$. This takes $5 	imes 3.8 = 19$ days.

Our target fraction is $\frac{1}{20}$, which sits comfortably between $\frac{1}{16}$ and $\frac{1}{32}$.

Therefore, the required time must strictly lie between **15.2 days** and **19 days**. Looking at the options, **16.5 days** is the only physically possible answer!

Animated Solution for Physics - Atoms and Nuclei: The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20th of the radon sample will remain undecayed, is (given $lo g_{10} e = 0.4343$ )

Select Answer:

Visualized Solution

$N = N_{0} e^{- λ t}$

$t_{1/2} = 3.8 days$

$N = \frac{N _{0}}{20}$

$ln (20) = λ t$

$λ = \frac{l n 2}{t _{1/2}}$

$t = 3.8 \times \frac{l o g _{10} 20}{l o g _{10} 2}$

$t \approx 16.42 days$

\text{Estimation Technique}

The Sigma Insight: Radioactivity

The Exponential Nature of Decay

Decoding the Half-Life

Setting Up the Target

The Power of Logarithms

The Final Calculation

The Pro-Tip

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=N0​e−λt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1/2​=3.8 days

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=20N0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(20)=λt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=t1/2​ln2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=3.8×log10​2log10​20​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t≈16.42 days