A radioactive sample is undergoing alpha-decay. At any time t1, its activity is A and another time t

Q: Why do we use $A = A_0 e^{-\lambda t}$ instead of $N = N_0 e^{-\lambda t}$?

Activity ($A$) is defined as the rate of decay, which is directly proportional to the number of active nuclei ($A = \lambda N$). Because of this direct proportionality, activity follows the exact same exponential decay law as the number of nuclei.

Q: What is the difference between average life and half-life?

Half-life ($T_{1/2}$) is the specific time it takes for exactly half of the sample to decay. Average life ($\tau$) is the statistical expected lifetime of any single nucleus before it decays. Mathematically, $\tau = 1/\lambda$ and $T_{1/2} = \ln(2)/\lambda$.

Q: Why did we divide the two equations?

We divided the equations to eliminate the unknown constant $A_0$. This is a standard algebraic trick in physics when dealing with proportional or exponential relationships, allowing us to isolate the relationship between the known variables.

Q: How does $\ln(e^x)$ simplify to $x$?

The natural logarithm ($\ln$) and the exponential function ($e$) are mathematical inverses of each other. Applying one undoes the effect of the other, which is why taking the natural log is the perfect tool to 'bring down' variables trapped in an exponent.

Q: Can the activity ever reach exactly zero?

Theoretically, an exponential decay curve approaches zero asymptotically, meaning it gets infinitely close but never truly reaches it. However, practically speaking, once all the discrete atoms in a finite sample have decayed, the activity becomes exactly zero.

Question

A radioactive sample is undergoing $\alpha$-decay. At any time $t_1$, its activity is $A$ and another time $t_2$, the activity is $\frac{A}{5}$. What is the average life time for the sample?

Options:

(A) $\frac{\ln 5}{t_2 - t_1}$

(B) $\frac{t_1 - t_2}{\ln 5}$

(C) $\frac{t_2 - t_1}{\ln 5}$

(D) $\frac{\ln(t_2 + t_1)}{2}$

Sigma Percentile · Accepted Answer

### Unraveling the Secrets of Radioactive Decay: Finding the Average Life

Imagine a radioactive sample as a bustling city of unstable atoms, each waiting for its moment to release energy and transform. This process isn't chaotic; it follows a beautifully strict mathematical rule known as the radioactive decay law. In this problem, we are going to act as nuclear detectives, using two snapshots in time to uncover the sample's average lifespan.

### The Mathematical Heartbeat of Decay

The activity of a radioactive sample, which tells us how many disintegrations occur per second, decays exponentially over time. We can express this fundamental behavior using the equation:

$$A(t) = A_0 e^{-\lambda t}$$

Here, $A_0$ represents the initial activity at the very beginning ($t=0$), and $\lambda$ is the decay constant—a unique fingerprint for every radioactive material that dictates how quickly it fades away.

### Two Snapshots in Time

Our problem gives us two specific clues. At a certain time $t_1$, the activity is measured to be $A$. We can plug this straight into our master equation:

$$A = A_0 e^{-\lambda t_1}$$

Later, at time $t_2$, the activity has dwindled down to $\frac{A}{5}$. Writing the equation for this second moment gives us:

$$\frac{A}{5} = A_0 e^{-\lambda t_2}$$

### The Power of Ratios

We now have a system of two equations, but we don't know the initial activity $A_0$. In physics, when you have an unknown constant multiplying your variables, taking a ratio is a powerful algebraic weapon. By dividing the first equation by the second, we can completely eliminate $A_0$:

$$\frac{A}{\frac{A}{5}} = \frac{A_0 e^{-\lambda t_1}}{A_0 e^{-\lambda t_2}}$$

On the left side, the $A$'s cancel out, leaving us with a neat $5$. On the right side, $A_0$ vanishes. Using the laws of exponents (subtracting the powers when dividing), we get:

$$5 = e^{\lambda(t_2 - t_1)}$$

### Logarithms to the Rescue

Our target, the decay constant $\lambda$, is trapped up in the exponent. To bring it down to ground level, we apply the natural logarithm ($\ln$) to both sides of the equation. Since $\ln(e^x) = x$, the exponential function is neutralized:

$$\ln 5 = \lambda(t_2 - t_1)$$

Rearranging this simple linear equation gives us the exact value of the decay constant:

$$\lambda = \frac{\ln 5}{t_2 - t_1}$$

### The Final Piece: Average Life

The question ultimately asks for the **average life time** of the sample. In nuclear physics, the average life (denoted by $	au$) is defined as the reciprocal of the decay constant. It represents the expected lifetime of any given nucleus before it decays.

$$	au = \frac{1}{\lambda}$$

By simply flipping our expression for $\lambda$, we arrive at our final, elegant answer:

$$	au = \frac{t_2 - t_1}{\ln 5}$$

This matches option (c). Understanding how to manipulate these exponential decay equations is a crucial skill for mastering modern physics!

Animated Solution for Physics - Atoms and Nuclei: A radioactive sample is undergoing $α$ -decay. At any time $t_{1}$ , its activity is $A$ and another time $t_{2}$ , the activity is $\frac{A}{5}$ . What is the average life time for the sample?

Select Answer:

Visualized Solution

Radioactive Decay Law

Activity at $t_{1}$ and $t_{2}$

Eliminating $A_{0}$

Simplifying the Ratio

Solving for Decay Constant $λ$

Average Life Time $τ$

The Way Forward

The Sigma Insight: Radioactivity

Unraveling the Secrets of Radioactive Decay

The Mathematical Heartbeat of Decay

Two Snapshots in Time

The Power of Ratios

Logarithms to the Rescue

The Final Piece

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