Problem 103 - A Complete Resource Book in Chemistry for JEE Main

Question

A sample of radioactive substance gave 630 counts per minute and 610 counts per minute at times differing by 1 hour. The decay constant \((\lambda)\) in \(\min ^{-1}\) is given by (a) \(\lambda=\frac{630}{610} \times 60\) (b) \(\mathrm{e}^{60 \mathrm{~A}}=\frac{630}{610}\) (c) \(\lambda=\frac{2.303}{60} \log \frac{610}{630}\) (d) \(\lambda=\frac{2.303}{60} \times \frac{630}{610}\)

Step-by-Step Solution

Answer

The decay constant is given by option (c) \( \lambda=\frac{2.303}{60} \log \frac{630}{610} \).

Step 1: Identify the decay formula

The decay of a radioactive substance over time can be described using the equation: \[ N(t) = N_0 e^{- ext{decay constant} imes t} \]where \( N(t) \) is the amount at time \( t \), and \( N_0 \) is the initial quantity. The decay constant can be expressed in terms of time in minutes when we know the initial and subsequent counts.

Step 2: Formulate the decay equation for this problem

Given that the counts reduce from 630 to 610 in 60 minutes, the decay equation becomes:\[ 610 = 630 e^{- ext{decay constant} imes 60} \]

Step 3: Solve the decay constant equation

Rearrange the equation to solve for the decay constant:1. Divide both sides by 630: \[ \frac{610}{630} = e^{- ext{decay constant} imes 60} \]2. Take the natural logarithm of both sides: \[ ext{ln}\left(\frac{610}{630}\right) = - ext{decay constant} imes 60 \]

Step 4: Isolate the decay constant

Solve for the decay constant:\[ ext{decay constant} = -rac{1}{60} imes ext{ln}\left(\frac{610}{630}\right) \]

Step 5: Convert natural logarithm to common logarithm

Apply the rule of logarithms to convert it to a base-10 logarithm using:\[ ext{ln}(x) = 2.303 imes ext{log}_{10}(x) \]Therefore:\[ ext{decay constant} = -rac{1}{60} imes 2.303 imes ext{log}_{10}\left(\frac{610}{630}\right) \]Simplifying further, it becomes:\[ ext{decay constant} = rac{2.303}{60} imes ext{log}_{10}\left(\frac{630}{610}\right) \]

Step 6: Match the derived formula with given options

The correct option corresponding to the decay constant formula we derived is (c) \( \lambda=\frac{2.303}{60} \log \frac{630}{610} \).

Key Concepts

Decay Constant

Imagine you have a shrinking snowball. Every minute, it shrinks at a predictable rate. The decay constant in radioactivity is similar, representing the rate at which a radioactive substance decays.
In mathematical terms, it shows how quickly the substance is losing particles. In the equation, \( N(t) = N_0 e^{-\lambda \times t} \), the decay constant, \( \lambda \), tells us how fast \( N \) reduces from \( N_0 \) over time \( t \).

• \( \lambda \) is the decay constant.
• Higher \( \lambda \) means faster decay.
• Units for \( \lambda \) often match the time unit used, like per minute (\( \min^{-1} \)).

Understanding \( \lambda \) helps us calculate how quickly a radioactive substance becomes less active. In the exercise, we saw that to find \( \lambda \), we rearranged the equation to isolate it, then used logarithms to solve.

Radioactivity Equation

The radioactivity equation is like a magical formula that lets us see into the future of decay. It helps predict how much of a substance remains after a certain time period.
This formula involves an exponential function because radioactivity diminishes over time, not disappearing all at once. The defining equation is:

• \( N(t) = N_0 e^{-\lambda \times t} \)
• \( N(t) \) is the remaining amount after time \( t \).
• \( N_0 \) is the original quantity.
• \( e \) is the base of the natural logarithm (approximately 2.718).

This exponential decay means the substance is halved over regular intervals, not by a fixed amount. In our exercise, the initial counts changed from 630 to 610 in one hour, demonstrating this gradual process.

Half-Life Computation

The half-life of a substance is the time it takes for half of it to decay. It's a handy concept because it gives a clear picture of how long a substance stays active. Mathematically, it's calculated using the decay constant. The relationship between half-life and the decay constant is:

• \( t_{1/2} = \frac{\ln(2)}{\lambda} \)
• Here, \( \ln(2) \) approximates to 0.693.

This calculation does not give exact durations always but gives a repetitive cycle duration for halving. In practical terms, if you know the decay constant, you can determine the half-life to predict how much substance will remain after several halves have passed.
In our exercise, deriving the decay constant led us one step closer to understanding this half-life, even though the computation focused on \( \lambda \) itself, our understanding of how things decay is based on this core concept.