A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 uF

Q: Why do we use the discharging equation for the capacitor?

The problem states the capacitor is 'first charged and later connected with resistor R'. This is the exact definition of a discharging RC circuit, where the initial charge $q_0$ decays through the resistor.

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

Half-life is the time required for half the sample to decay ($t_{1/2} = \ln(2)/\lambda$), while average life is the expected lifetime of a single nucleus ($\tau = 1/\lambda$). The problem specifically gives the average life.

Q: Why must the exponential terms be equal?

If you have a ratio like $e^{-at} / e^{-bt}$, it simplifies to $e^{(b-a)t}$. For this to be a constant value independent of time $t$, the exponent must be zero, which means $b-a = 0$, or $a=b$.

Q: Can I use the activity formula $A = \lambda N$ directly?

Yes! The activity $A(t)$ is simply $\lambda N(t)$. Since $N(t) = N_0 e^{-\lambda t}$, the activity is $A(t) = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$. The exponential time dependence is exactly the same.

Q: Why did we convert milliseconds and microfarads to standard units?

In physics, standard SI units (seconds, Farads, Ohms) ensure that the final calculated value is in the correct standard unit (Ohms). Mixing prefixes like milli and micro without converting them will lead to massive calculation errors.

Question

A radioactive sample has an average life of $30	ext{ ms}$ and is decaying. A capacitor of capacitance $200\ \mu	ext{F}$ is first charged and later connected with resistor $R$. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time, then the value of $R$ should be ......... $\Omega$.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in the unexpected connections between seemingly unrelated phenomena. In this problem, we are asked to bridge the gap between the macroscopic world of electromagnetism (an RC circuit) and the microscopic quantum world of nuclear physics (a radioactive sample).

### Analyzing the Setup

Imagine two completely different physical systems side by side. On the left, we have a capacitor that has been fully charged and is now discharging through a resistor. On the right, we have a sample of radioactive material quietly decaying away.

The problem gives us a fascinating constraint: the ratio of the charge on the capacitor, $q(t)$, to the activity of the radioactive sample, $A(t)$, is perfectly constant over time.

To understand what this means, we must first write down the equations governing both systems. For a discharging capacitor, the charge decays exponentially according to the formula:
$$q(t) = q_0 e^{-\frac{t}{RC}}$$
where $RC$ is the time constant of the circuit.

Similarly, the activity of a radioactive sample (which is the rate of decay) also follows an exponential decay law:
$$A(t) = A_0 e^{-\lambda t}$$
where $\lambda$ is the decay constant of the material.

### The Master Equation

Now, let's substitute these expressions into our given condition. We are told that the ratio $\frac{q(t)}{A(t)}$ is constant.
$$\frac{q(t)}{A(t)} = \frac{q_0 e^{-\frac{t}{RC}}}{A_0 e^{-\lambda t}}$$

For this ratio to be completely independent of time $t$, the exponential terms must perfectly cancel each other out. If they didn't, the ratio would either grow or shrink as time went on. Therefore, the exponents must be identical:
$$e^{-\frac{t}{RC}} = e^{-\lambda t}$$

Taking the natural logarithm of both sides, we get:
$$-\frac{t}{RC} = -\lambda t$$
$$\lambda = \frac{1}{RC}$$

This is our master equation! It elegantly states that the decay constant of the radioactive sample must equal the reciprocal of the RC time constant.

### Final Calculation

We need to find the resistance $R$. Rearranging our master equation, we get:
$$R = \frac{1}{\lambda C}$$

But wait, what is $\frac{1}{\lambda}$? In nuclear physics, the reciprocal of the decay constant is exactly the **average life** ($	au$) of the radioactive sample!
$$R = \frac{	au}{C}$$

Now, we simply plug in the given values. The average life $	au$ is $30	ext{ ms}$, and the capacitance $C$ is $200\ \mu	ext{F}$. We must be extremely careful to convert these into standard SI units to avoid silly mistakes.
$$	au = 30 	imes 10^{-3}	ext{ s}$$
$$C = 200 	imes 10^{-6}	ext{ F}$$

Substituting these into our equation:
$$R = \frac{30 	imes 10^{-3}}{200 	imes 10^{-6}}$$
$$R = \frac{30}{200} 	imes 10^3$$
$$R = 0.15 	imes 1000 = 150\ \Omega$$

The required resistance is exactly **$150\ \Omega$**. This problem is a brilliant reminder that the mathematics of exponential decay is a universal language, describing everything from the flow of electrons to the splitting of atomic nuclei!

Enter Numerical Value:

Visualized Solution

$q (t)$ and $A (t)$

Exponential Decay Laws

Equating the Exponentials

Relating $λ$ and $R C$

Solving for $R$

Substituting Values

Final Calculation

The Way Forward

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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.

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........ .

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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

In given time $t = 0$ , Activity of two radioactive substances $A$ and $B$ are equal. After time $t$ , the ratio of their activities $\frac{R _{B}}{R _{A}}$ decreases according to $e^{- 3 t}$ . If the half life of $A$ is $ln 2$ , the half-life of $B$ will be

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

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q(t) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(t)

Exponential Decay Laws

Equating the Exponentials

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

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

Substituting Values

Final Calculation

The Way Forward

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A radioactive sample is undergoing α-decay. At any time t1​, its activity is A and another time t2​, the activity is 5A​. What is the average life time for the sample?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The half-life of 215At is 100μs. The time taken for the activity of a sample of 215At to decay to 161​th of its initial value is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The activity of a radioactive sample falls from 700 s−1 to 500 s−1 in 30 min. Its half-life is close to

$q (t)$ and $A (t)$

Relating $λ$ and $R C$

Solving for $R$

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?

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

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