In a sample initially there are equal number of atoms of two radioactive isotopes A and B. 3 days later the number of atoms of A is twice that of B. Half life of B is `1.5` days. What is half life of isotope A? (in days)

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Initial Conditions
Let the initial number of atoms of both isotopes A and B be \( N_0 \). Therefore, we have:
- Initial number of atoms of A, \( N_A(0) = N_0 \)
- Initial number of atoms of B, \( N_B(0) = N_0 \)

### Step 2: Write the Decay Equations
The number of atoms remaining after a certain time \( t \) can be expressed using the half-life formula:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{\tau}} \]
where \( \tau \) is the half-life of the isotope.

For isotope A:
\[ N_A(3) = N_0 \left( \frac{1}{2} \right)^{\frac{3}{\tau_A}} \]

For isotope B (with a half-life of \( \tau_B = 1.5 \) days):
\[ N_B(3) = N_0 \left( \frac{1}{2} \right)^{\frac{3}{1.5}} = N_0 \left( \frac{1}{2} \right)^{2} = N_0 \cdot \frac{1}{4} \]

### Step 3: Relate the Number of Atoms After 3 Days
According to the problem, after 3 days, the number of atoms of A is twice that of B:
\[ N_A(3) = 2 \cdot N_B(3) \]
Substituting the expressions we derived:
\[ N_0 \left( \frac{1}{2} \right)^{\frac{3}{\tau_A}} = 2 \cdot \left( N_0 \cdot \frac{1}{4} \right) \]
This simplifies to:
\[ N_0 \left( \frac{1}{2} \right)^{\frac{3}{\tau_A}} = \frac{N_0}{2} \]

### Step 4: Cancel \( N_0 \) and Simplify
Assuming \( N_0 \neq 0 \), we can divide both sides by \( N_0 \):
\[ \left( \frac{1}{2} \right)^{\frac{3}{\tau_A}} = \frac{1}{2} \]

### Step 5: Equate the Exponents
Since the bases are the same, we can equate the exponents:
\[ \frac{3}{\tau_A} = 1 \]

### Step 6: Solve for \( \tau_A \)
Rearranging gives:
\[ \tau_A = 3 \text{ days} \]

### Final Answer
The half-life of isotope A is \( \tau_A = 3 \) days.

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