Problem 39 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

During the nuclear explosion, one of the products is \({ }^{90} \mathrm{Sr}\) with half life of \(6.93\) years. If \(1 \square \mathrm{g}\) of \({ }^{90} \mathrm{Sr}\) was absorbed in the bones of a newly born baby in place of \(\mathrm{Ca}\), how much time, in years, is required to reduce it by \(90 \%\) if it is not lost metabolically. [Main Jan. 07, 2020 (I)]

Step-by-Step Solution

Answer

It takes approximately 23 years for the amount of \(^{90}\mathrm{Sr}\) to reduce by 90%.

Step 1: Understand the Problem

We need to find out how long it takes for the amount of \(^{90}\mathrm{Sr}\) to decrease by 90%, which means only 10% remains. This relates to radioactive decay, involving half-life.

Step 2: Use the Half-life Formula

The formula to calculate the remaining quantity \(N\) of a radioactive material after time \(t\) is given by: \( N = N_0 \times \left(\frac{1}{2}\right)^{t/T} \), where \(N_0\) is the initial quantity and \(T\) is the half-life.

Step 3: Set Up the Equation

We know \(N_0 = 1\) gram and we want \(N/N_0 = 0.1\) or 10% remaining. Replacing in the equation: \(0.1 = \left(\frac{1}{2}\right)^{t/T}\).

Step 4: Solve for Time \(t\)

Take the logarithm of both sides: \(\log_{10}(0.1) = t \cdot \log_{10}(1/2) / T\). Simplify \(\log_{10}(0.1) = -1\) and \(\log_{10}(1/2) \approx -0.3010\). Substitute to solve \(t = -1 \cdot T / (-0.3010)\).

Step 5: Calculate \(t\) with Given Half-life

Substitute \(T = 6.93\) years: \(t = -1 \cdot 6.93 / -0.3010 \approx 23\) years.

Key Concepts

Half-life

Half-life is a concept in nuclear chemistry that describes the time it takes for half of a given quantity of a radioactive substance to decay.
This is an exponential decay process, meaning that it does not decrease linearly with time. Instead, the amount of the substance reduces by half in each half-life period.For example, if you start with a certain amount of a radioactive isotope, after one half-life, half of it remains. After two half-lives, a quarter remains, and so on, following the formula: \[ N = N_0 \times \left(\frac{1}{2}\right)^{t/T} \]where:

• \(N\) = remaining quantity
• \(N_0\) = initial quantity
• \(t\) = time elapsed
• \(T\) = half-life

Understanding half-life is vital when analyzing radioactive decay problems. It helps predict how long a radioactive substance remains active.

Radioisotope

A radioisotope, also known as a radioactive isotope, is an atom that has excess nuclear energy, making it unstable.
This instability leads to the emission of radiation as the atom attempts to reach a stable state.
Radioisotopes are commonly used in medical diagnostics, treatment, and research.Some important points about radioisotopes include:

• They can occur naturally or be artificially created.
• Commonly used isotopes include Iodine-131 for thyroid diagnosis and treatment, and Carbon-14 for radiocarbon dating.
• The radiation emitted can be in forms such as alpha particles, beta particles, or gamma rays.

In the context of the exercise, Strontium-90 (\(^{90}\mathrm{Sr}\)) is a radioisotope produced in nuclear reactions, and understanding its decay is crucial during calculations involving radioactive materials.

Nuclear Chemistry

Nuclear chemistry is the branch of chemistry that deals with the study of changes in atomic nuclei.
Unlike conventional chemistry, which involves electron cloud interactions, nuclear chemistry focuses on reactions that alter the nucleus of an atom.
The changes usually involve the release or absorption of energy. Key concepts in nuclear chemistry include:

• Radioactive decay: the process by which an unstable atomic nucleus loses energy by emitting radiation.
• Nuclear reactions: reactions that alter the composition, structure, and energy of atomic nuclei.
• Applications involve power generation, nuclear medicine, and industrial uses.

In solving problems like the one provided, understanding nuclear decay equations and half-life calculations is essential. It allows scientists to measure the rate of decay and predict how much of a radioisotope will be present at a given time. This is critical for both safety considerations and practical applications.