JEE Advanced 2023

LEVELJEE Main

Animated Solution for Physics - Atoms and Nuclei: In a radioactive decay process, the activity is defined as $A = - \frac{d N}{d t}$ , where $N (t)$ is the number of radioactive nuclei at time $t$ . Two radioactive sources, $S_{1}$ and $S_{2}$ have same activity at time $t = 0$ . At a later time, the activities of $S_{1}$ and $S_{2}$ are $A_{1}$ and $A_{2}$ , respectively. When $S_{1}$ and $S_{2}$ have just completed their $3^{rd}$ and $7^{th}$ half-lives, respectively, the ratio $A_{1} / A_{2}$ is __________.

Q: Why did we use the formula with $(1/2)^n$ instead of $e^{-\lambda t}$?

Both formulas are mathematically equivalent since $n = t/T_{1/2}$ and $T_{1/2} = \ln(2)/\lambda$. However, when the time is given directly in terms of the number of half-lives ($n$), using $(1/2)^n$ is much faster and avoids unnecessary calculations with $e$ and $\ln(2)$.

Q: What does it mean physically that $S_2$ has completed 7 half-lives while $S_1$ has only completed 3?

It means that $S_2$ decays much faster than $S_1$. In the exact same amount of time $t$, $S_2$ has halved its activity 7 times, whereas $S_1$ has only halved it 3 times. Therefore, $S_2$ has a much shorter half-life.

Q: Can the number of half-lives be a fraction or a decimal?

Absolutely! The formula $A = A_0(1/2)^n$ works for any real number $n$. For example, if a sample has undergone 2.5 half-lives, its activity would be $A_0(1/2)^{2.5}$.

Q: Why did the ratio $A_1/A_2$ turn out to be greater than 1?

Because $S_1$ has undergone fewer half-lives (3) compared to $S_2$ (7). This means $S_1$ has retained a larger fraction of its initial activity, making $A_1$ larger than $A_2$.

Enter Numerical Value:

Visualized Solution

$A_{1} (0) = A_{2} (0) = A_{0}$

$A_{1} (0) = A_{2} (0) = A_{0}$

$A (t) = A_{0} (\frac{1}{2})^{n}$

$A (t) = A_{0} (\frac{1}{2})^{n}$
$n = \frac{t}{T _{1/2}}$

$A_{1} and A_{2}$

$A_{1} = A_{0} (\frac{1}{2})^{3}$
$A_{2} = A_{0} (\frac{1}{2})^{7}$

$\frac{A _{1}}{A _{2}}$

$\frac{A _{1}}{A _{2}} = \frac{A _{0} ( \frac{1}{2} ) ^{3}}{A _{0} ( \frac{1}{2} ) ^{7}}$

$Final Calculation$

$\frac{A _{1}}{A _{2}} = (\frac{1}{2})^{3 - 7}$
$= (\frac{1}{2})^{- 4}$
$= 2^{4} = 16$

$T_{1/2} = \frac{l n 2}{λ}$

$T_{1/2} = \frac{l n 2}{λ}$

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The Sigma Insight: Radioactivity

Solution Diagram

Analyzing the Setup

Imagine two radioactive samples,

S_{1}

and

S_{2}

, sitting on a lab bench. At time

t = 0

, they are equally active, meaning they are emitting the exact same number of particles per second. We call this initial activity

A_{0}

Now, how does activity change over time? The law of radioactive decay tells us that after

n

half-lives, the activity drops by a factor of

2^{n}

. So, the activity

A

is simply

A_{0}

times

(1/2)^{n}

The Master Equation

Let's look at a specific later time

t

. The problem states that

S_{1}

has completed exactly 3 half-lives, while

S_{2}

has completed 7 half-lives in the same duration. This means

S_{2}

is decaying much faster!

We can write their new activities,

A_{1}

and

A_{2}

, by substituting

n = 3

and

n = 7

into our formula. For the first source, we get

A_{1} = A_{0} (1/2)^{3}

. For the second source, we get

A_{2} = A_{0} (1/2)^{7}

Final Calculation

We need the ratio of

A_{1}

A_{2}

. Let's divide the two expressions. Notice how the initial activity,

A_{0}

, beautifully cancels out from the numerator and the denominator. We are left with

(1/2)^{3}

divided by

(1/2)^{7}

Using the laws of exponents,

3 - 7

gives us

- 4

. So we have

(1/2)^{- 4}

, which is simply

2^{4}

. Evaluating this, we get 16.

So,

S_{1}

is 16 times more active than

S_{2}

at this moment. This was a straightforward application of the half-life formula. Always remember, the number of half-lives doesn't have to be an integer, but when it is, the calculations are incredibly fast!

Common Traps & FAQs

Why did we use the formula with

(1/2)^{n}

instead of

e^{- λ t}

Both formulas are mathematically equivalent since

n = t / T_{1/2}

and

T_{1/2} = ln (2) / λ

. However, when the time is given directly in terms of the number of half-lives (

n

), using

(1/2)^{n}

is much faster and avoids unnecessary calculations with

e

and

ln (2)

Does the initial number of nuclei

N_{0}

have to be the same for both sources?

Not necessarily! The problem states that their initial *activities* (

A_{0}

) are the same. Since

A_{0} = λ N_{0}

, if their decay constants (

λ

) are different, their initial number of nuclei (

N_{0}

) must also be different to yield the same

A_{0}

What does it mean physically that

S_{2}

has completed 7 half-lives while

S_{1}

has only completed 3?

It means that

S_{2}

decays much faster than

S_{1}

. In the exact same amount of time

t

S_{2}

has halved its activity 7 times, whereas

S_{1}

has only halved it 3 times. Therefore,

S_{2}

has a much shorter half-life.

Can the number of half-lives be a fraction or a decimal?

Absolutely! The formula

A = A_{0} (1/2)^{n}

works for any real number

n

. For example, if a sample has undergone 2.5 half-lives, its activity would be

A_{0} (1/2)^{2.5}

Why did the ratio

A_{1} / A_{2}

turn out to be greater than 1?

Because

S_{1}

has undergone fewer half-lives (3) compared to

S_{2}

(7). This means

S_{1}

has retained a larger fraction of its initial activity, making

A_{1}

larger than

A_{2}

Enter Numerical Value:

Visualized Solution

$A_{1} (0) = A_{2} (0) = A_{0}$

$A (t) = A_{0} (\frac{1}{2})^{n}$

$A_{1} and A_{2}$

$\frac{A _{1}}{A _{2}}$

$Final Calculation$

$T_{1/2} = \frac{l n 2}{λ}$

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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

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Two radioactive substances $A$ and $B$ have decay constants $5 λ$ and $λ$ , respectively. At $t = 0$ , a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(\frac{1}{e})^{2}$ will be

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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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(t)=A0​(21​)n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​ and A2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​A1​​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1/2​=λln2​

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

$A_{1} (0) = A_{2} (0) = A_{0}$

$A (t) = A_{0} (\frac{1}{2})^{n}$

$A_{1} and A_{2}$

$\frac{A _{1}}{A _{2}}$

$Final Calculation$

$T_{1/2} = \frac{l n 2}{λ}$