Question

How much time is required for a \(6.25\) -mg sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75 \mathrm{mg}\) if it has a half-life of \(27.8\) days?

Short answer

It will take approximately \(77.6\) days for the \(6.25\) mg sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75\) mg.

Step-by-step solution

Understand the radioactive decay formula

The formula for radioactive decay is given by \(A(t) = A_0 e^{-\lambda t}\), where \(A(t)\) is the amount of the substance remaining after time \(t\), \(A_0\) is the initial amount, \(\lambda\) is the decay constant, and \(e\) is the base of natural logarithm.

Calculate the decay constant using the half-life

To calculate the decay constant, we'll use the formula: \(\lambda = \frac{\ln{2}}{T_{1/2}}\), where \(T_{1/2}\) is the half-life. For \({ }^{51} \mathrm{Cr}\), the half-life is \)27.8$ days, so the decay constant will be: \(\lambda = \frac{\ln{2}}{27.8} \)

Plug in the initial and final amounts into the radioactive decay formula

We know that the initial amount \(A_0\) is \(6.25\) mg, and the final amount \(A(t)\) is \(0.75\) mg. Plugging in these values into the radioactive decay formula, we get: \(0.75 = 6.25 e^{-\frac{\ln{2}}{27.8}t}\)

Solve for time \(t\)

To solve for \(t\), we'll first divide both sides by \(6.25\) mg: \[\frac{0.75}{6.25} = e^{-\frac{\ln{2}}{27.8}t}\] Next, we'll take the natural logarithm of both sides: \[\ln{\left(\frac{0.75}{6.25}\right)} = -\frac{\ln{2}}{27.8}t\] Finally, isolate \(t\) by dividing both sides by \(-\frac{\ln{2}}{27.8}\): \[t = \frac{\ln{\left(\frac{0.75}{6.25}\right)}}{-\frac{\ln{2}}{27.8}}\]

Calculate the time

Now, we'll use a calculator to find \(t\): \[t = \frac{\ln{\left(\frac{3}{25}\right)}}{-\frac{\ln{2}}{27.8}} \approx 77.6 \ \text{days}\] It will take approximately \(77.6\) days for the \(6.25\) mg sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75\) mg.

Key concepts

Half-Life

Understanding half-life is key to grasping how radioactive substances decay over time. Simply put, half-life is the amount of time it takes for half of a radioactive substance to lose its radioactivity. This concept reveals the rate at which unstable atoms undergo decay, turning them into more stable forms.

To illustrate, if a sample has a half-life of 10 years, it means that after 10 years, only half of the original amount of the radioactive material will remain; the other half will have decayed. With another 10 years, half of the remaining material will decay, leaving a quarter of the original amount. This process continues exponentially, following a predictable pattern that helps scientists determine the age of materials and predict the decay of radioactive substances.

Half-life is an essential concept when dealing with radioactive materials, as it determines their longevity and potential impact on the environment or human health. It is also a fundamental concept in archeological dating methods like radiocarbon dating.

Decay Constant

The decay constant, often represented by the symbol \(\lambda\), is a probability measure of the likelihood of decay of a radioactive isotope. It is inversely related to half-life—essentially, the decay constant tells us how quickly a radioactive material will decay. For any radioisotope, the decay constant is a fixed value and remains unaffected by external conditions such as temperature or pressure.

To calculate the decay constant when the half-life is known, the following formula is used: \(\lambda = \frac{\ln{2}}{T_{1/2}}\), where \(T_{1/2}\) represents the half-life of the substance. \(\ln{2}\) is used here because it is the natural logarithm of 2, which correlates with the fact that half-life is the time taken for half of the substance to decay. It's a direct window into the atomic nature of matter, demonstrating how atoms deteriorate at a predictable rate that can be modeled mathematically.

Natural Logarithm

The natural logarithm, often denoted as \(\ln\), is a logarithmic function with the base \(e\) (Euler's number, approximately equal to 2.71828). This special logarithm is widely used in science and engineering because of its natural appearance in growth and decay problems, including radioactive decay calculations.

When dealing with exponential decay equations, taking the natural logarithm of both sides of an equation is a common step to solve for an unknown variable. This is because the natural logarithm is the inverse operation of exponentiation with base \(e\). Therefore, it allows us to move from the exponential form to a linear one, which is much easier to manipulate algebraically. This quality makes it indispensable for solving time-related decay problems, as seen in the example of radioactive decay in this exercise.

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. This pattern is commonly observed in nature and is especially relevant to radioactive decay. The formula \(A(t) = A_0 e^{-\lambda t}\) succinctly captures the essence of exponential decay; it shows that the amount of a radioactive substance \(A(t)\) at any given time \(t\) reduces from its original amount \(A_0\) by a factor depending on the decay constant \(\lambda\) and the base of natural logarithms \(e\).

The 'exponential' aspect refers to the presence of the term \(e^{-\lambda t}\) in the decay equation, indicating that the decrease is not linear but rather diminishes more rapidly at first and then slows down as time goes on. Such behavior is typical with radioactive substances where the rate of decay is fastest initially and gradually reduces as fewer radioactive atoms are left to decay. Thus, understanding exponential decay is fundamental to interpreting how radioactive materials change over time.