Problem 100 - A Complete Resource Book in Chemistry for JEE Main

Question

What weight of \(\mathrm{C}^{14}\) will have radioactivity one curie if \(\lambda\) (disintegration constant) is \(4.4 \times 10^{-12} \mathrm{sec}^{-1} ?\) (a) \(3.7 \times 10^{-6} \mathrm{~kg}\) (b) \(51 \times 10^{-3} \mathrm{~kg}\) (c) \(1.96 \times 10^{-4} \mathrm{~kg}\) (d) \(1.7 \times 10^{-6} \mathrm{~kg}\)

Step-by-Step Solution

Answer

The weight is \(1.96 \times 10^{-4}\) kg, corresponding to option (c).

Step 1: Understand the Problem

We need to find the mass of a radioactive isotope, \( \mathrm{C}^{14} \), that results in a radioactivity of one curie. The disintegration constant \( \lambda \) is given.

Step 2: Relate Activity to Curie

One curie is defined as \(3.7 \times 10^{10}\) disintegrations per second. This means that the activity \(A\) for one curie is \(3.7 \times 10^{10}\) disintegrations per second.

Step 3: Use the Activity Formula

The activity \(A\) of a radioactive sample is given by \(A = \lambda N\), where \(\lambda\) is the disintegration constant and \(N\) is the number of radioactive atoms.

Step 4: Relate to Number of Atoms

Since \(A = 3.7 \times 10^{10}\) and \(\lambda = 4.4 \times 10^{-12}\), we can find \(N\) by rearranging the formula: \(N = \frac{A}{\lambda}\). Substitute the known values: \(N = \frac{3.7 \times 10^{10}}{4.4 \times 10^{-12}}.\)

Step 5: Calculate the Number of Atoms

Calculate \(N = \frac{3.7 \times 10^{10}}{4.4 \times 10^{-12}} = 8.41 \times 10^{21}\) atoms.

Step 6: Convert Atoms to Mass

We know 1 mole of \(\mathrm{C}^{14}\) contains Avogadro's number \( = 6.022 \times 10^{23} \) atoms, and its molar mass is 14 grams. Therefore, mass \(m\) can be calculated as: \(m = \left(\frac{8.41 \times 10^{21}}{6.022 \times 10^{23}}\right) \times 14. \)

Step 7: Final Calculation

Compute the mass: \(m = 1.96 \times 10^{-4} \text{ kg}. \)

Step 8: Identify the Correct Option

The calculated mass \(1.96 \times 10^{-4}\) kg matches option (c).

Key Concepts

Disintegration Constant

The disintegration constant, denoted as \( \lambda \), is a fundamental concept in the study of radioactive decay. It represents the probability of an individual nucleus of a radioactive substance to decay per unit of time. In simpler terms, it tells us how rapidly or slowly a radioactive isotope breaks down. For example, if \( \lambda \) is large, the substance decays quickly, resulting in a shorter half-life. If it's small, the decay process is slower, and the substance has a longer half-life. This constant is expressed in units of time\(^{-1}\), for instance, \(\mathrm{sec}^{-1}\) or \(\mathrm{year}^{-1}\). Thus, the given disintegration constant for \( \mathrm{C}^{14} \) is \( 4.4 \times 10^{-12} \mathrm{sec}^{-1} \), indicating a slow decay process characteristic of \( \mathrm{C}^{14} \). Understanding \( \lambda \) helps us calculate the activity of a radioactive isotope at any given moment.

Activity Formula

Activity \( A \) refers to the rate at which a radioactive sample undergoes decay, which means it tells us how many nuclear disintegrations occur per second. In nuclear physics, the activity formula is expressed as \( A = \lambda N \).Here:

• \( A \) is the activity measured in disintegrations per second,
• \( \lambda \) is the disintegration constant, and
• \( N \) is the number of atoms of the radioactive isotope.

This relationship is crucial as it connects the disintegration constant to the measurable activity of the sample.For instance, if you know the activity (such as the given \( 3.7 \times 10^{10} \) disintegrations per second which equals one curie), and the disintegration constant, you can rearrange to solve for \( N \): \( N = \frac{A}{\lambda} \). This allows you to determine how many radioactive atoms are present.

Radioactive Isotope Mass Calculation

When dealing with radioactive isotopes, calculating the total mass present from the number of atoms is a fundamental task. After determining the number of atoms \( N \) using the activity and disintegration constant, the next step is to find the mass \( m \). This is achieved by connecting \( N \) to moles and using the molar mass of the isotope. The steps are as follows:

• Identify Avogadro's number: \( 6.022 \times 10^{23} \) atoms/mole.
• Know the molar mass of the isotope. For \( \mathrm{C}^{14} \), it's 14 g/mole.

Now, mass \( m \) can be calculated using:\[m = \left(\frac{N}{6.022 \times 10^{23}}\right) \times 14. \]By substituting \( N \) we selected earlier, we find the total mass of \( \mathrm{C}^{14} \) in our sample. This computation results in a mass of \( 1.96 \times 10^{-4} \text{ kg} \).

Curie

A curie is an older, but still frequently used unit to measure radioactive activity. It marks a specific amount of radioactivity or decay events that occur per second. One curie is precisely defined as \( 3.7 \times 10^{10} \) disintegrations per second. Named after the pioneering scientists Marie and Pierre Curie, this unit helps scientists communicate and measure large quantities of radioactivity in simple and effective terms.Though now largely replaced by the becquerel (where 1 curie = 37 billion becquerels), the curie remains in practical use, particularly in the United States, due to its historical significance. Understanding curies helps contextualize radioactive decay rates and how we measure them across various isotopes and applications.