Problem 34 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An experiment requires minimum beta activity product at the rate of 346 beta particles per minute. The half life period of \({ }_{42}^{99} \mathrm{Mo}\), which is a beta emitter is \(66.6\) hours. Find the minimum amount of \({ }_{42}^{99}\) Mo required to carry out the experiment in \(6.909\) hours.

Step-by-Step Solution

Answer

Approximately 372 moles of \\(^ {99}_{42} Mo\\) are required.

Step 1: Understand the Problem

We need the minimum amount of \(^ {99}_{42} Mo\) that will provide at least 346 beta particles per minute, given its half-life is 66.6 hours.

Step 2: Use Decay Formula

The decay formula is \((N_t = N_0 e^{-\lambda t})\), where \(N_t\) is the remaining quantity after time \(t\), \(N_0\) is the initial quantity, and \(\lambda\) is the decay constant.

Step 3: Calculate the Decay Constant

The decay constant \(\lambda\) is calculated using the formula \((\lambda = \frac{0.693}{T_{1/2}})\), where \(T_{1/2}\) is the half-life. For \(T_{1/2} = 66.6\) hours, \(\lambda = \frac{0.693}{66.6} = 0.0104\) hr\(^{-1}\).

Step 4: Relate Beta Activity to Number of Atoms

The rate of beta emission is proportional to the number of undecayed atoms. So, Activity, \(A = \lambda N_t = 346\) particles/minute at time t = 6.909 hours.

Step 5: Calculate Initial Amount

Using the relationship found in previous steps, \(\lambda N_0 = 346 e^{-0.0104 \times 6.909}\). Solve for \(N_0\) to find the initial quantity needed.\[N_0 = \frac{346} {0.0104} e^{0.0104 \times 6.909} \approx 346 \times 1.07406 \approx 371.66\] Therefore, approximately 372 moles of \(^ {99}_{42} Mo\) are needed to maintain the activity.

Key Concepts

Beta Activity

Beta activity is a term used to describe the rate at which a radioactive substance emits beta particles. Beta particles can be high-energy, high-speed electrons or positrons emitted from certain types of radioactive decay, known as beta decay. When an unstable atom undergoes beta decay, it changes into another element by converting a neutron into a proton (or vice versa), releasing a beta particle in the process. To understand how beta activity is calculated, it's helpful to grasp that activity is generally measured in particles per unit of time, such as particles per minute or per second. In the context of our problem, we are interested in maintaining beta activity at a rate of 346 beta particles per minute. Key points about beta activity:

• It depends on the number of radioactive atoms present and their decay rate.
• The higher the number of atoms, the higher the beta activity, assuming the decay rate remains constant.

Recognizing the relationship between beta activity, the number of atoms, and the decay constant is crucial for solving problems related to radioactive substances.

Half-life Period

The half-life period of a radioactive isotope is the time it takes for half of its unstable atoms to decay into something else. For example, if you start with 1000 beta-emitting atoms, after one half-life, you'd expect only 500 of those original atoms to remain undecayed.The concept of half-life is essential in radioactive decay as it provides a measure of how quickly a substance decays. It varies widely among different isotopes, ranging from fractions of a second to billions of years. In our exercise, the half-life of \( ^{99}_{42} ext{Mo} \, \) is given as 66.6 hours.Understanding half-life is crucial because:

• It helps predict how long a radioactive substance will remain active.
• It provides data needed to calculate the decay constant, which links directly to the rate of beta activity.

In any calculations involving radioactive decay, the half-life serves as a fundamental constant that helps predict the number of atoms remaining at any given time.

Decay Constant

The decay constant, symbolized by \( \lambda \, \), is a probability rate constant that provides information about the likelihood of a single atom decaying per unit time. It is an essential parameter in nuclear physics because it helps quantify the intensity of radioactive decay.The decay constant is inherently related to the half-life of a radioactive substance. They are linked by the formula: \( \lambda = \frac{0.693}{T_{1/2}} \) where \( T_{1/2} \) is the half-life of the substance.Why is the decay constant important?

• It quantifies the speed of the decay process, offering insight into how quickly the substance will lose its radioactivity.
• It allows us to calculate the expected remaining quantity of a substance over time using the exponential decay formula \( (N_t = N_0 e^{-\lambda t}) \).

In our given problem, the calculation of the decay constant leads to the determination of how much of the beta emitter \( ^{99}_{42} \text{Mo} \, \) is needed to achieve the desired beta activity during the experiment.