Question

The half-life of a radioactive substance is 2 days. Find the time required for a given amount of the material to decay to \(1 / 10\) of its original mass.

Short answer

Answer: The time required for the radioactive substance to decay to 1/10 of its original mass is approximately 3.32 days.

Step-by-step solution

Determine the decay constant (\(\lambda\)) using the half-life

The formula that relates half-life, T, with the decay constant, \(\lambda\), is given by: \(T = \cfrac{\ln 2}{\lambda}\). We are given the half-life, T = 2 days. We can solve this equation for the decay constant, \(\lambda\), as follows: \(\lambda = \cfrac{\ln 2}{T}\). Now, plug in the value of half-life, T = 2 days: \(\lambda = \cfrac{\ln 2}{2}\).

Determine the time required for the substance to decay to 1/10 of its original mass

We will now use the exponential decay formula: \(N(t) = N_0 e^{-\lambda t}\). We need to find \(t\) when \(N(t) = \cfrac{1}{10} N_0\). This is the condition when the substance has decayed to 1/10 of its original mass. So, we have: \(\cfrac{1}{10} N_0 = N_0 e^{- \cfrac{\ln 2}{2} t}\). We can cancel out \(N_0\) from both sides and simplify to get: \(\cfrac{1}{10} = e^{-\cfrac{\ln 2}{2} t}\). Now, to solve for \(t\), we can take the natural logarithm of both sides: \(\ln \cfrac{1}{10} = -\cfrac{\ln 2}{2} t\). Now, we can isolate \(t\) by dividing both sides by \((-\cfrac{\ln 2}{2})\): \(t = \cfrac{\ln{\frac{1}{10}}}{-\cfrac{\ln 2}{2}}\). And finally, we can simplify: \(t = \cfrac{2\ln{\frac{1}{10}}}{-\ln 2}\). Now, we can calculate the value of \(t\): \(t \approx 3.32\) days.

Interpret the result

The time required for the given amount of the radioactive substance to decay to 1/10 of its original mass is approximately 3.32 days.

Key concepts

Half-Life

The concept of half-life is a fundamental measure in radioactive decay that helps describe how quickly a substance decreases in mass over time. Half-life, denoted as \( T \), is the period it takes for a radioactive material to reduce to half of its initial quantity. Imagine you start with a pile of radioactive marbles; the half-life tells us how long it'll take for half of them to change or decay. Understanding half-life is key to solving problems related to radioactive decay, as it provides a direct relationship between time and decay rate. Whether dealing with atoms in a substance or understanding the longevity of cosmic elements, half-life offers an invaluable piece of the puzzle.

Decay Constant

The decay constant \( \lambda \) is another crucial component when understanding radioactive decay. It represents the probability of decay per unit time. By knowing the decay constant, we can describe how fast a substance breaks down. The relationship between half-life and the decay constant is expressed with this formula:

• \( T = \frac{\ln 2}{\lambda} \)

Where \( \ln 2 \) is the natural logarithm of 2, approximately equal to 0.693. For example, in our exercise with a half-life of 2 days, we solve for \( \lambda \) as follows:

• \( \lambda = \frac{\ln 2}{2} \)

This shows that the decay constant helps link the concept of decay to the time it takes for decay processes to occur.

Exponential Decay Formula

The exponential decay formula is a mathematical expression used to model the process of substances decreasing over time. The formula is:

• \( N(t) = N_0 e^{-\lambda t} \)

Here, \( N(t) \) represents the amount of substance remaining at time \( t \), \( N_0 \) is the initial amount of the substance, and \( \lambda \) is the decay constant.This formula shows how the quantity of the material decreases exponentially, which means it reduces by a constant proportion in equal time intervals. So, the smaller the decay constant, the slower the decay rate. In our problem, we see how this formula gets adjusted when \( N(t) \) is one-tenth of \( N_0 \). We set:

• \( \frac{1}{10} N_0 = N_0 e^{-\frac{\ln 2}{2} t} \)

By simplifying this, we can find out exactly how much time it takes for a known change in mass.

Natural Logarithm

Natural logarithms, denoted as \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.718 \). They simplify the process of solving equations involving exponential terms, often arising within scientific and engineering problems, such as radioactive decay.In our radioactive decay exercise, natural logarithms allow us to isolate time in the exponential decay formula. For instance, when solving \( \frac{1}{10} = e^{-\frac{\ln 2}{2} t} \), we use the natural logarithm to transform it into:

• \( \ln \frac{1}{10} = -\frac{\ln 2}{2} t \)

This makes it easier to calculate \( t \) by rearranging the equation to find:

• \( t = \frac{2\ln{\frac{1}{10}}}{-\ln 2} \)

By understanding natural logarithms, we are better equipped to handle complex decay rate calculations, providing clarity and precision in results.