Problem 91 - Master Resource Book in JEE Main Physics

Question

A source contains two phosphorous radio nuclides \({ }_{15}^{32} \mathrm{P}\left(T_{1 / 2}=14.3\right.\) days) and \({ }_{15}^{33} \mathrm{P}\left(T_{1 / 2}=25.3\right.\) days). Initially, \(10 \%\) of the decay come from \({ }_{15}^{33} \mathrm{P}\). How long one must wait until \(90 \%\) do so? (a) 250 days (b) 295 days (c) 305 days (d) 208 days

Step-by-Step Solution

Answer

295 days

Step 1: Understand the Types of Decay

We have two different nuclides: \( _{15}^{32}\mathrm{P} \) and \( _{15}^{33}\mathrm{P} \), each with different half-lives. Initially, \(10\% \) of the decay comes from \( _{15}^{33}\mathrm{P}\). The goal is to find the time when \(90\%\) of the decay will come from \( _{15}^{33}\mathrm{P}\).

Step 2: Define the Rate of Decay

The decay rate \( N(t) \) of a nuclide is given by \( N(t) = N_0 \cdot e^{-\lambda t} \), where \( \lambda = \frac{\ln(2)}{T_{1/2}} \), \( N_0 \) is the initial amount, and \( T_{1/2} \) is the half-life. Calculate \( \lambda_{32} \) and \( \lambda_{33} \) for \( _{15}^{32}\mathrm{P} \) and \( _{15}^{33}\mathrm{P} \) respectively.

Step 3: Calculate Decay Constants

For \( _{15}^{32}\mathrm{P} \), the decay constant \( \lambda_{32} = \frac{\ln(2)}{14.3} \). For \( _{15}^{33}\mathrm{P} \), the decay constant \( \lambda_{33} = \frac{\ln(2)}{25.3} \). These constants will define how quickly each nuclide decays.

Step 4: Set Up the Ratio Equation

Initially, the decay rates are such that \( \frac{R_{33}(0)}{R_{32}(0) + R_{33}(0)} = 0.10 \). After waiting time \( t \), this ratio should change to \( \frac{R_{33}(t)}{R_{32}(t) + R_{33}(t)} = 0.90 \). The decay rates \( R(t) \) are given by \( R_{32}(t) = k_{32} N_{32} e^{-\lambda_{32} t} \) and \( R_{33}(t) = k_{33} N_{33} e^{-\lambda_{33} t} \).

Step 5: Solve for Time \( t \)

Set up the ratio equation with exponential terms: \[ \frac{k_{33} N_{33} e^{-\lambda_{33} t}}{k_{32} N_{32} e^{-\lambda_{32} t} + k_{33} N_{33} e^{-\lambda_{33} t}} = 0.90 \]. Simplify and solve this equation for \( t \). Note that constants \( k_{32} \) and \( k_{33} \) cancel out, as do \( N_{32} \) and \( N_{33} \) if initially equal.

Step 6: Simplify and Calculate

Simplifying the equation using \( R_{33}(t)/R_{32}(t) = 9 \), leads to \( e^{-(\lambda_{33} - \lambda_{32}) t} = 9 \). Solving gives \[ t = \frac{\ln(9)}{\lambda_{32} - \lambda_{33}} \]. Substitute \( \lambda_{32} \) and \( \lambda_{33} \) to find \( t \approx 295 \) days.

Key Concepts

Half-life calculation

Half-life is a fundamental concept in the study of radioactive decay. It is the time required for half of the radioactive nuclei in a sample to decay. This means that after one half-life period, only 50% of the original nuclei remain. Calculating the half-life of a radioactive substance helps in determining how it dissipates over time, which is crucial for various applications, from radiometric dating to medical treatments.

To calculate the half-life, we use the formula:\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Here, \( T_{1/2} \) is the half-life, and \( \lambda \) represents the decay constant. This formula indicates the inverse relationship between the half-life and the decay constant—when one increases, the other decreases. Understanding half-life assists in making predictions about the longevity and potency of radioactive materials.

Decay constant

The decay constant \( \lambda \) reflects the probability of decay of a radioactive nuclide per unit time. It is a crucial parameter because it determines how quickly a radioactive material will decay. A higher \( \lambda \) implies a faster rate of decay, leading to a shorter half-life.

The formula to determine \( \lambda \) from the half-life is:\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
By utilizing the natural logarithm of 2, this expression offers a direct relationship between the decay constant and the half-life. Calculating \( \lambda \) helps scientists and engineers understand the stability of a radioactive nuclide and predict its behavior in different conditions.

• For \( _{15}^{32}\mathrm{P} \), \( \lambda = \frac{\ln(2)}{14.3} \) days.
• For \( _{15}^{33}\mathrm{P} \), \( \lambda = \frac{\ln(2)}{25.3} \) days.

In this context, these constants allow us to compare and understand the relative decay speeds of \( _{15}^{32}\mathrm{P} \) and \( _{15}^{33}\mathrm{P} \). The decay constant plays a pivotal role in setting up the equations necessary to analyze radioactive decay scenarios.

Radioactive nuclides

Radioactive nuclides, also known as radioisotopes, are unstable atoms that decay over time. They emit radiation in the form of alpha, beta, or gamma particles. This process of emission is called radioactive decay.

Each nuclide is characterized by its unique decay properties, including half-life and decay constant, which dictate how it transforms. In the original exercise, the focus is on two specific phosphorus isotopes: \( _{15}^{32}\mathrm{P} \) and \( _{15}^{33}\mathrm{P} \). Each isotope has its own half-life and decay constant, affecting how quickly it loses its radioactive properties.

• \( _{15}^{32}\mathrm{P} \) has a half-life of 14.3 days.
• \( _{15}^{33}\mathrm{P} \) has a half-life of 25.3 days.

Understanding these features is essential for practical purposes, including health technologies, archaeological dating, and energy production. Harnessing the decay characteristics of these nuclides allows for strategic applications in diverse scientific fields.