Question

A radioactive isotope decays in such a way that the number of atoms present at a given time, \(N(t)\), obeys the equation $$ \frac{d N}{d t}=-\lambda N $$ If there are initially \(N_{0}\) atoms present, find \(N(t)\) at later times.

Short answer

The number of atoms at time \( t \) is given by \( N(t) = N_0 e^{-\lambda t} \).

Step-by-step solution

- Identify the differential equation

The given differential equation is \( \frac{d N}{d t} = -\lambda N \). This equation tells us that the rate of decay of the number of atoms is proportional to the number of atoms present.

- Separate the variables

Rewrite the equation so that all terms involving \(N\) are on one side and all terms involving \(t\) are on the other side: Let's separate the variables: \( \frac{1}{N} \frac{d N}{d t} = -\lambda \).

- Integrate both sides

Integrate both sides to find \(N(t)\). \( \int \frac{1}{N} dN = -\lambda \int dt \). The left side integrates to \( \ln|N| \), and the right side integrates to \( -\lambda t \). So, \( \ln|N| = -\lambda t + C \), where \(C\) is the constant of integration.

- Solve for \(N(t)\)

Exponentiate both sides to solve for \(N(t)\): \( N = e^{-\lambda t + C} \). We can rewrite \(e^C\) as another constant, say \(C_1\). So, \( N = C_1 e^{-\lambda t} \).

- Apply the initial condition

Use the initial condition that when \( t = 0 \), \( N = N_0 \): \( N_0 = C_1 e^0 \). This means \( C_1 = N_0 \).

- Write the final expression for \( N(t) \)

Substitute \( C_1 \) back into the equation: \( N(t) = N_0 e^{-\lambda t} \). This is the expression for the number of atoms at any later time \( t \).

Key concepts

Exponential Decay

Radioactive decay follows a pattern known as exponential decay. This means the quantity of radioactive atoms decreases at a rate proportional to the number of atoms remaining. If you start with a certain number of atoms, their reduction over time can be modeled with the equation: \(\frac{dN}{dt} = -\text{λ}N\).
Here, \(N(t)\) is the number of atoms at time \(t\), and \(\text{λ}\) (lambda) is the decay constant. This constant is unique to each radioactive material and determines how quickly the atoms decay.
The exponential decay formula derived from the solution, \(N(t) = N_0 e^{-\text{λ}t}\), shows how the number of atoms decreases over time. This equation tells us that the number of atoms decreases exponentially, becoming a specific fraction of the original amount each second, year, or other time unit.

Separation of Variables

To solve the differential equation, we use a technique called separation of variables. This involves rearranging the equation so that each variable appears on one side of the equation.
Starting from \(\frac{dN}{dt} = -\text{λ}N\), we separate the variables by moving all terms involving \(N\) to one side and those involving \(t\) to the other: \(\frac{1}{N} \frac{dN}{dt} = -\text{λ}\).
Now, we can integrate each side independently. This method simplifies the problem, making it easier to find a solution for \(N(t)\).
Remember, the goal is to isolate the variables so we can use integration to solve it.

Initial Conditions

Initial conditions are essential in solving differential equations uniquely. In this case, the initial condition is given as \(N(0) = N_0\), meaning that at time \(t = 0\), the number of atoms is \(N_0\).
Applying the initial conditions helps us determine the value of the arbitrary constant in the solution.
After integrating and solving, we introduce the initial condition to find the specific solution to our problem. For example, after integrating, we get \(\text{ln}|N| = -\text{λ}t + C\). Using the initial condition, we solve for \(C\) and integrate it into our final equation, \(N(t) = N_0 e^{-\text{λ}t}\).
This process ensures our solution accurately reflects the initial state of the system.

Integration

Integration is a core part of solving differential equations. It allows us to reverse differentiation to find the original function.
In our problem, we reached the integral form: \(\text{∫}\frac{1}{N} dN = -\text{λ} ∫dt\).
The left side integrates to \(\text{ln}|N|\), and the right side to \(-\text{λ}t\). Thus, we get: \(\text{ln}|N| = -\text{λ}t + C\).
By exponentiating both sides, we solve for \(N(t)\): \(N = e^{-\text{λ}t + C} = C_1 e^{-\text{λ}t}\).
In this step, \(e^C\) is written as a new constant \(C_1\). Applying the initial condition, as seen, we get \(N(t) = N_0 e^{-\text{λ}t}\). Integration plays a crucial role in transforming differential equations into a solvable function for our specific conditions.