A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10^3 disintegrations pe

Q: Why didn't we use the initial activity of $10^3$ disintegrations per second?

The fraction of nuclei that decay depends only on the decay constant and the time elapsed. The initial activity is extra information that isn't needed for this specific calculation.

Q: Can I solve this without using the approximation?

Yes, you could calculate $1 - e^{-0.04}$ directly if you had a scientific calculator. However, in exams like JEE, the approximation $e^{-x} \approx 1 - x$ for small $x$ is expected and makes the calculation much faster.

Q: Where does the approximation $e^{-x} \approx 1 - x$ come from?

It comes from the Maclaurin series expansion of $e^{-x}$, which is $1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$. When $x$ is very small, the terms with $x^2$ and higher powers become negligible.

Q: Why do we calculate $1 - e^{-\lambda t}$ instead of just $e^{-\lambda t}$?

The term $e^{-\lambda t}$ gives the fraction of nuclei that remain *undecayed*. The question specifically asks for the fraction that *will decay*, which is the complement, $1 - e^{-\lambda t}$.

Q: What are the units of the decay constant $\lambda$?

The decay constant has units of inverse time. Since the half-life is given in seconds, $\lambda$ will be in $\text{s}^{-1}$.

Question

A freshly prepared sample of a radioisotope of half-life $1386	ext{ s}$ has activity $10^3$ disintegrations per second. Given that $\ln 2 = 0.693$, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first $80	ext{ s}$ after preparation of the sample is

Sigma Percentile · Accepted Answer

### The Ticking Clock of Radioactivity

Imagine a freshly prepared sample of a radioisotope as a room full of ticking clocks, each set to go off at a random time. While we can't predict when a specific clock will ring, we know exactly how the entire room behaves over time. This predictable behavior is governed by the **law of radioactive decay**.

In this problem, we are given a sample with a half-life of $1386	ext{ s}$ and an initial activity of $10^3$ disintegrations per second. Our goal is to find the percentage of nuclei that decay within the first $80	ext{ s}$.

### Decoding the Decay Constant

The first step in any radioactivity problem is often finding the decay constant, $\lambda$. This constant tells us the probability of decay per unit time. It is intimately connected to the half-life, $T_{1/2}$, through the relation:

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

We are given $\ln 2 = 0.693$ and $T_{1/2} = 1386	ext{ s}$. Plugging these values in, we get:

$$ \lambda = \frac{0.693}{1386} = 5 	imes 10^{-4} 	ext{ s}^{-1} $$

This tiny number means that in any given second, a very small fraction of the nuclei will decay.

### The Fraction of Decayed Nuclei

The number of active nuclei remaining at time $t$ is given by the exponential decay law:

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

However, we are interested in the number of *decayed* nuclei, $N_d$. This is simply the initial number minus the remaining number:

$$ N_d = N_0 - N(t) = N_0(1 - e^{-\lambda t}) $$

To find the fraction of decayed nuclei, we divide by $N_0$:

$$ 	ext{Fraction decayed} = \frac{N_d}{N_0} = 1 - e^{-\lambda t} $$

### The Power of Approximation

Now, we need to evaluate this fraction at $t = 80	ext{ s}$. Let's first calculate the exponent $\lambda t$:

$$ \lambda t = (5 	imes 10^{-4} 	ext{ s}^{-1})(80	ext{ s}) = 0.04 $$

Since $0.04$ is much less than $1$, we can use a powerful mathematical tool: the **Taylor series approximation**. For very small values of $x$, the exponential function $e^{-x}$ can be approximated as $1 - x$.

Applying this to our fraction:

$$ 1 - e^{-\lambda t} \approx 1 - (1 - \lambda t) = \lambda t $$

This brilliant shortcut saves us from calculating complex exponentials!

### The Final Percentage

Using our approximation, the fraction of decayed nuclei is simply equal to $\lambda t$, which is $0.04$.

To express this as a percentage, we multiply by $100$:

$$ 	ext{Percentage decayed} = 0.04 	imes 100 = 4\% $$

And there we have it! In the first $80	ext{ s}$, exactly $4\%$ of the initial nuclei will have decayed. Notice how the initial activity of $10^3$ disintegrations per second was a red herring—we didn't need it at all to find the fraction!

Enter Numerical Value:

Visualized Solution

$Visualizing Radioactive Decay$

$Fraction of Decayed Nuclei$

$Decay Constant$

$Calculating λ$

$Value of λ$

$Evaluating λ t$

$Small x Approximation$

$Final Percentage$

The Sigma Insight: Radioactivity

The Ticking Clock of Radioactivity

Decoding the Decay Constant

The Fraction of Decayed Nuclei

The Power of Approximation

The Final Percentage

Similar Questions

There are $1 0^{10}$ radioactive nuclei in a given radioactive element. Its half-life time is $1 min$ . How many nuclei will remain after $30 s$ ? ( $2 = 1.414$ )

The radioactive decay rate of a radioactive element is found to be $1 0^{3}$ disintegration/second at a certain time. If the half-life of the element is one second, the decay rate after one second is ....... and after three seconds is........ .

The activity of a radioactive sample falls from $700 s^{- 1}$ to $500 s^{- 1}$ in $30 min$ . Its half-life is close to

In a radioactive material, fraction of active material remaining after time $t$ is $\frac{9}{16}$ . The fraction that was remaining after time $t /2$ is

The activity of a freshly prepared radioactive sample is $1 0^{10}$ disintegrations per second, whose mean life is $1 0^{9}$ s. The mass of an atom of this radioisotope is $1 0^{- 25}$ kg. The mass (in mg) of the radioactive sample is

At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei, (b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number.

A radioactive sample $S_{1}$ having an activity of $5 μ Ci$ has twice the number of nuclei as another sample $S_{2}$ which has an activity of $10 μ Ci$ . The half lives of $S_{1}$ and $S_{2}$ can be

The half-life of $^{215} At$ is $100 μ s$ . The time taken for the activity of a sample of $^{215} At$ to decay to $\frac{1}{16} th$ of its initial value is

Half-lives of two radioactive elements $A$ and $B$ are $20 min$ and $40 min$ , respectively. Initially, the samples have equal number of nuclei. After $80 min$ , the ratio of decayed numbers of $A$ and $B$ nuclei will be

Half lives of two radioactive nuclei $A$ and $B$ are $10$ minutes and $20$ minutes, respectively. If initially a sample has equal number of nuclei, then after $60$ minutes, the ratio of decayed numbers of nuclei $A$ and $B$ will be

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing Radioactive Decay

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fraction of Decayed Nuclei

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decay Constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Value of λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating λt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Small x Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Percentage