Question

The disintegration rate for a sample containing \(_{27}^{60} \mathrm{Co}\) as the only radioactive nuclide is 6740 dis \(\mathrm{h}^{-1}\). The half-life of 20 Co is 5.2 years. Estimate the number of atoms of \(_{27}^{60}\) Co in the sample.

Short answer

Therefore, we estimate the number of atoms of \(_{27}^{60}\) Co in the sample to be approximately \(4.4 x 10^{8}\) atoms.

Step-by-step solution

Identify Given Information

From the problem, we know that the disintegration rate (R) is 6740 disintegrations per hour, and the half-life (t1/2) of \(_{27}^{60}\) Co is 5.2 years.

Convert half-life to seconds

Since R is given in disintegrations per hour, it will be beneficial to convert the t1/2 from years to seconds. There are 8760 hours in a year, so \(t1/2 = 5.2 * 8760 = 45552\) hours.

Calculate Decay Constant

The decay constant (\( \lambda \)) is related to the half-life through the equation \( t1/2 = \frac{0.693}{\lambda} \). Solving for \( \lambda \) we get \( \lambda = \frac{0.693}{t1/2} \) which equals \(\frac{0.693}{45552} = 1.52 x 10^{-5} h^{-1} \).

Determine the Number of Atoms

The disintegration rate is given by R = \( \lambda*N \), where N is the number of radioactive atoms. Solving for N, we get \( N = \frac{R}{\lambda} \) which equals \( \frac{6740}{1.52 x 10^{-5}} = 4.4 x 10^{8} \) atoms.

Key concepts

Decay Constant

The decay constant, often denoted by the symbol \(\lambda\), is a crucial concept in understanding radioactive decay. The decay constant represents the probability per unit time that a given atom will decay. It is a measure of how quickly a radioactive substance undergoes decay.

The mathematical relationship between the decay constant and half-life is given by the formula:\[ t_{1/2} = \frac{0.693}{\lambda} \]

Here, \(t_{1/2}\) stands for the half-life of the substance. This formula highlights that the decay constant is inversely related to the half-life. A larger decay constant indicates a shorter half-life, meaning the substance decays more quickly.

In the original problem, the decay constant for \(_{27}^{60} \mathrm{Co}\) was calculated using this relationship and found to be \(1.52 \times 10^{-5} \ \mathrm{h}^{-1}\). Understanding the decay constant is essential for calculating how many atoms of a radioactive nuclide are present in a sample by relating it to the disintegration rate.

Half-Life

The half-life of a radioactive substance is the time it takes for half of the sample's radioactive atoms to decay. It is a fundamental concept that helps us understand how quickly a substance undergoes decay.

The half-life is independent of the initial amount of the substance, and it is a constant for any given nuclide. This means that regardless of how much radioactive material you start with, the half-life remains the same.

For \(_{27}^{60} \mathrm{Co}\), the half-life is 5.2 years, as given in the exercise. This translates to a significant duration over which the radioactive decay occurs.

Knowing the half-life allows us to predict how a substance will decay over time and is crucial for applications such as carbon dating and nuclear medicine.

Disintegration Rate

The disintegration rate, often denoted as \(R\), refers to the number of decay events occurring per unit of time in a sample of radioactive material. It provides a measurable way to quantify the activity of a radioactive sample.

The relationship between the disintegration rate and the number of radioactive atoms \(N\) is described by the equation:\[ R = \lambda \times N \]

Where \(\lambda\) is the decay constant. This equation allows us to calculate the number of radioactive atoms present in a sample, given the disintegration rate and the decay constant.

In the exercise, the disintegration rate was given as 6740 disintegrations per hour. By rearranging the equation, the number of atoms \(N\) was determined, highlighting the connection between these key concepts in radioactive decay.