Question

A radioactive substance decay \(25 \%\) in 10 minute. If at start there are \(4 \times 10^{20}\) atoms present, after what time will the number of atoms be reduced to \(10^{20}\) atoms? (given \(\ln 3=1.098\) ) (a) \(10.98 \mathrm{~min}\) (b) \(21.97 \mathrm{~min}\) (c) \(48.19 \mathrm{~min}\) (d) None of these

Short answer

The time t when the number of atoms will be reduced to 1 * 10^{20} is approximately 10.98 minutes (Option a).

Step-by-step solution

Understand the Decay Process

First, recognize that radioactive decay is an exponential process and can be described by the formula N(t) = N_0 * e^(-kt), where N(t) is the number of atoms at time t, N_0 is the original number of atoms, k is the decay constant, and t is the time elapsed. We need to find the time t when N(t) = 1 * 10^{20} atoms.

Calculate the Decay Constant

Using the information that 25% of the substance decays in 10 minutes, we have N(t) = 0.75N_0 after 10 minutes. Plugging this into the decay formula gives us 0.75N_0 = N_0 * e^(-10k). Solving for k, we get k = -(1/10) * (ln 0.75).

Find the Time for the Desired Decay

Now we need to find t such that N(t) = N_0 * e^(-kt) = 1 * 10^{20}. Because we know that N_0 = 4 * 10^{20}, we can substitute N_0 and k into the decay formula and solve for t.

Calculate the Natural Logarithm of the Ratio N(t)/N_0

Since we want N(t) = 1 * 10^{20} and N_0 = 4 * 10^{20}, the ratio N(t)/N_0 = 1/4. The natural logarithm of this ratio is ln(1/4).

Solve for Time

By substituting ln(1/4) = -ln(4) into the equation ln(N(t)/N_0) = -kt, and using our calculated value of k, we can solve for the time t when N(t) will be 1 * 10^{20}.

Substitute Known Values and Solve for t

Plugging in the known values, including the given natural logarithm ln(3), we solve for t using the relationship ln(4) = ln(3) + ln(1.3333), where ln(1.3333) can be approximated using the half-life information.

Key concepts

Exponential Decay

Exponential decay is a fundamental concept in understanding radioactive decay and involves a quantity decreasing at a rate proportional to its current value. This concept plays a key role in various fields including chemistry, physics, and even finance. When dealing with radioactive substances, or other scenarios where decay is proportional to the amount remaining, the mathematical representation of exponential decay becomes vital.

In mathematical terms, if you have a substance that decays over time, the amount of substance left, denoted as N(t), after time t can be modeled by the equation
\[ N(t) = N_0 \cdot e^{(-kt)} \]
where \( N_0 \) is the initial amount of the substance and \( k \) is the decay constant, unique to each decaying element. The base of the exponential function, \( e \), is the natural number approximately equal to 2.71828, which arises naturally in various growth and decay processes.

Decay Constant

The decay constant, represented by the symbol \( k \), is the cornerstone of quantifying radioactive decay. It signifies the probability of a single atom decaying per unit time and is specific to each radioactive isotope. The larger the value of the decay constant, the more quickly the substance decays.

To find the decay constant from a given percentage of decay over time, we use the initial decay equation and rearrange it to solve for \( k \). For example, if 25% of a substance decays in 10 minutes, the remaining amount after 10 minutes will be 75% (or 0.75 times) of the initial amount. Thus, the equation becomes \( 0.75N_0 = N_0 \cdot e^{(-10k)} \), and solving for \( k \) gives us the decay constant specific to the substance in question.

Half-Life Calculation

The half-life of a radioactive substance is the time required for half of the radioactive atoms in a sample to decay. This concept is pivotal in radioactive dating techniques and managing nuclear waste. To calculate half-life from the decay constant, we use the relationship
\[ t_{1/2} = \frac{\ln(2)}{k} \]
where \( t_{1/2} \) is the half-life and \( k \) is the decay constant. It provides a convenient way to discuss the long-term behavior of radioactive substances without dealing with the complexity of the entire decay process. In some cases, knowledge of a substance's half-life can inform us about the rate at which a substance decays, even without knowing the decay constant explicitly.

Natural Logarithm in Decay

The natural logarithm arises naturally in the context of exponential growth and decay processes because it is the inverse operation to taking an exponent with base \( e \). In the case of radioactive decay, we commonly use the natural logarithm to manipulate the decay equation to solve for the time or the decay constant.

By taking the natural logarithm of both sides of the decay equation, we can linearize the equation. For example, for the ratio \( N(t)/N_0 = 1/4 \), the natural logarithm helps us find the time, t, at which the original amount of substance is reduced to one-fourth, which is represented mathematically as \( \ln(1/4) \). Understanding how to use the natural logarithm function is essential in solving decay problems.

Solving Decay Equations

Solving decay equations involves isolating the variable of interest, typically the time \( t \), and using the decay formula. Needed are the initial amount \( N_0 \), the final amount \( N(t) \), and the decay constant \( k \). To find the time when the substance is reduced to a certain quantity, the steps often include calculating the decay constant, using the natural logarithm to linearize the equation, and then solving for the time.

As an example, with an initial quantity of \( 4 \times 10^{20} \) atoms, a final quantity of \( 1 \times 10^{20} \) atoms, and a decay constant derived from known decay percentage over time, we would insert these values into the exponential decay formula, employ the properties of logarithms to solve for t, and obtain the time at which the initial quantity decays to the desired amount.