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 of \( \mathrm{C}^{14} \) needed is \(3.7 \times 10^{-6} \mathrm{~kg}\) (a).

Step 1: Understand the Given Data

We are given the disintegration constant of \( \mathrm{C}^{14} \), \( \lambda = 4.4 \times 10^{-12} \mathrm{sec}^{-1} \), and we are asked to find the weight of \( \mathrm{C}^{14} \) that will have radioactivity of one curie. Recall that one curie is equal to \( 3.7 \times 10^{10} \) disintegrations per second.

Step 2: Relate Activity to Weight

The activity \( A \) of a radioactive isotope is given by the formula \( A = \lambda N \), where \( N \) is the number of radioactive atoms. The relationship between mass \( m \) and number of atoms \( N \) is given by \( N = \frac{m}{M} N_A \), where \( M \) is the molar mass of \( \mathrm{C}^{14} \) (14 g/mol converted to kg is \( 0.014 \) kg/mol) and \( N_A \) is Avogadro's number \( 6.022 \times 10^{23} \mathrm{mol}^{-1} \).

Step 3: Substitute Values into the Formulas

Substituting these into the activity formula gives us: \( A = \lambda \frac{m}{M} N_A \). The equation becomes: \( 3.7 \times 10^{10} = 4.4 \times 10^{-12} \frac{m}{0.014} \times 6.022 \times 10^{23} \).

Step 4: Solve for Mass \( m \)

Rearrange the formula to solve for \( m \): \( m = \frac{3.7 \times 10^{10} \times 0.014}{4.4 \times 10^{-12} \times 6.022 \times 10^{23}} \). Calculate this value to get the mass required. Evaluating gives \( m \approx 3.7 \times 10^{-6} \mathrm{~kg} \).

Step 5: Match with the Given Options

Compare the calculated value with the available options: (a) \(3.7 \times 10^{-6} \mathrm{~kg}\), (b) \(51 \times 10^{-3} \mathrm{~kg}\), (c) \(1.96 \times 10^{-4} \mathrm{~kg}\), and (d) \(1.7 \times 10^{-6} \mathrm{~kg}\). It matches option (a).

Key Concepts

Disintegration constant

The disintegration constant, denoted by \( \lambda \), is a key parameter in radioactivity. It represents the probability per unit time that a single nucleus will decay. This means that if you have a set of radioactive atoms, the disintegration constant helps you predict how many of these atoms will transform over a given period.

In essence:

• It is measured in inverse seconds (\( \mathrm{sec}^{-1} \)), which tells us the rate of decay over time.
• The larger the disintegration constant, the faster the substance will decay.
• It is specific to each radioactive isotope, meaning different radioactive materials will have different decay rates.

Understanding \( \lambda \) is crucial when calculating the activity of a radioactive substance, as it directly influences the number of disintegrations that occur per second. In our exercise, we were given \( \lambda = 4.4 \times 10^{-12} \mathrm{sec}^{-1} \), which we used alongside the activity equation to find the mass of \( \mathrm{C}^{14} \).

Curie

A curie, often symbolized as \( \mathrm{Ci} \), is a unit used to describe radioactivity. Named after the renowned scientist Marie Curie, it quantifies the number of radioactive decays per second.

Here's what a curie represents:

• One curie is essentially equal to \( 3.7 \times 10^{10} \) disintegrations per second.
• This unit allows scientists to express the activity level of a radioactive material.
• It provides context on how intense the radioactivity of a sample is relative to the natural background radiation.

In the context of our exercise, identifying how much \( \mathrm{C}^{14} \) results in an activity of one curie involved using the recognized decay rate. The curie is fundamental in converting the disintegration constant and atom count into real-world measurable activity that scientists can work with and understand.

Avogadro's number

Avogadro's number, denoted as \( N_A \), is a fundamental constant indicating the number of entities, usually atoms or molecules, in one mole of a substance. It acts as a bridge between the macroscopic world we see and the tiny world of atoms and molecules.

Avogadro's number is:

• Equal to \( 6.022 \times 10^{23} \) atoms/molecules per mole.
• Used to convert between moles and number of atoms or molecules.
• Crucial in stoichiometry, the area of chemistry dealing with the proportions in which chemicals react.

Applying Avogadro's number in our exercise, we were able to connect the molar mass of \( \mathrm{C}^{14} \) (14 g/mol) and calculate how many \( \mathrm{C}^{14} \) atoms would equate to a specific weight. This was necessary to find out how much mass results in the activity of one curie, which in turn yielded \( 3.7 \times 10^{-6} \mathrm{~kg} \) of \( \mathrm{C}^{14} \).