Two radioactive materials X1 and X2 have decay constants 10λ and λ respectively. If initially they h

Q: Why do we assume the initial number of nuclei is the same?

The problem explicitly states 'If initially they have the same number of nuclei', which allows us to use the same $N_0$ for both and cancel it out in the ratio.

Q: What does a larger decay constant mean physically?

A larger decay constant (like $10\lambda$ for $X_1$) means the material decays much faster compared to a material with a smaller decay constant (like $\lambda$ for $X_2$).

Q: Why does the ratio $N_1/N_2$ decrease over time?

Since $X_1$ decays faster than $X_2$, the number of remaining nuclei $N_1$ drops more rapidly than $N_2$, causing their ratio to decrease from 1 (at $t=0$) to smaller fractions.

Q: Can we use half-life instead of the decay constant?

Yes, but it would add unnecessary steps. The decay constant $\lambda$ is directly given, and the ratio involves $e$, making the exponential decay formula $N = N_0 e^{-\lambda t}$ the most direct approach.

Q: What if the initial number of nuclei were different?

If they started with different amounts, say $N_{01}$ and $N_{02}$, the ratio would be $\frac{N_{01}}{N_{02}} e^{-9\lambda t} = \frac{1}{e}$, and the time $t$ would depend on their initial ratio.

Question

Two radioactive materials $X_1$ and $X_2$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $1/e$ after a time

Options:

(A) $\frac{1}{10\lambda}$

(B) $\frac{1}{11\lambda}$

(C) $\frac{11}{10\lambda}$

(D) $\frac{1}{9\lambda}$

Sigma Percentile · Accepted Answer

### The Setup

Imagine you have two different radioactive materials, $X_1$ and $X_2$. They both start with the exact same number of unstable nuclei, which we will call $N_0$. However, they don't decay at the same rate. The problem tells us that the decay constant for $X_1$ is $10\lambda$, while the decay constant for $X_2$ is just $\lambda$.

What does this mean physically? A larger decay constant means the material is highly unstable and decays very rapidly. So, $X_1$ is decaying ten times faster than $X_2$. As time goes on, the number of remaining nuclei for $X_1$ will drop much faster than for $X_2$.

### The Master Equation

To solve this, we need our trusty radioactive decay law. The number of undecayed nuclei $N(t)$ at any given time $t$ is given by the exponential decay formula:

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

Let's write this out specifically for our two materials. For $X_1$, the decay constant is $10\lambda$, so its equation is:

$$N_1(t) = N_0 e^{-10\lambda t}$$

For $X_2$, the decay constant is simply $\lambda$, so its equation is:

$$N_2(t) = N_0 e^{-\lambda t}$$

### The Mathematical Execution

The question asks us to find the specific time $t$ when the ratio of the remaining nuclei of $X_1$ to $X_2$ becomes exactly $1/e$. Let's set up this mathematical condition:

$$\frac{N_1(t)}{N_2(t)} = \frac{1}{e}$$

Now, we substitute the expressions we found earlier into this ratio:

$$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e}$$

Notice something beautiful here? The initial number of nuclei, $N_0$, is present in both the numerator and the denominator. This means it completely cancels out! The exact starting amount doesn't matter, only the ratio of their decay rates.

### The Final Takeaway

After canceling $N_0$, we are left with an equation involving exponents of $e$:

$$\frac{e^{-10\lambda t}}{e^{-\lambda t}} = e^{-1}$$

Using the fundamental laws of exponents, when we divide terms with the same base, we subtract their powers. So, $-10\lambda t - (-\lambda t)$ becomes $-9\lambda t$. Our equation simplifies beautifully to:

$$e^{-9\lambda t} = e^{-1}$$

Since the bases on both sides are identical (both are $e$), their exponents must be equal. We can directly equate them:

$$-9\lambda t = -1$$

Solving for $t$, we get our final, elegant answer:

$$t = \frac{1}{9\lambda}$$

And there you have it! By simply applying the decay law and using basic algebra, we found the exact moment when the ratio hits $1/e$.

Animated Solution for Physics - Atoms and Nuclei: Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $10 λ$ and $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $1/ e$ after a time

Select Answer:

Visualized Solution

\text{Visualizing the Decay}

\text{Radioactive Decay Law}

\text{Equations for } X_1 \text{ and } X_2

\text{Setting up the Ratio}

\text{Substituting the Expressions}

\text{Simplifying the Exponents}

\text{Solving for } t

The Sigma Insight: Radioactivity

The Setup

The Master Equation

The Mathematical Execution

The Final Takeaway

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