The relation between half`-`life period `(T_(1//2))` and disintegration constant `(lambda)` is expressed as

Options

A. `T_(1//2) = (ln2)/(lambda)`
B. `T_(1//2) ln2 = lambda`
C. `T_(1//2) = (1)/(lambda)`
D. `(lambda + T_(1//2)) = (ln)/(2)`

Answer

To derive the relation between the half-life period \( T_{1/2} \) and the disintegration constant \( \lambda \), we can follow these steps:

### Step 1: Understand the Definition of Half-Life
The half-life \( T_{1/2} \) of a radioactive substance is defined as the time required for the quantity of the substance to reduce to half of its initial amount.

### Step 2: Use the Disintegration Formula
The number of radioactive atoms remaining after time \( t \) can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where:
- \( N(t) \) is the number of atoms remaining at time \( t \),
- \( N_0 \) is the initial number of atoms,
- \( \lambda \) is the disintegration constant,
- \( t \) is the time elapsed.

### Step 3: Set Up the Equation for Half-Life
At half-life \( T_{1/2} \), the number of atoms remaining is half of the initial amount:
\[ N(T_{1/2}) = \frac{N_0}{2} \]
Substituting this into the disintegration formula gives:
\[ \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \]

### Step 4: Simplify the Equation
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\lambda T_{1/2}} \]

### Step 5: Take the Natural Logarithm
To solve for \( T_{1/2} \), take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\lambda T_{1/2} \]

### Step 6: Use the Property of Logarithms
Using the property of logarithms, we can rewrite \( \ln\left(\frac{1}{2}\right) \) as:
\[ \ln\left(\frac{1}{2}\right) = -\ln(2) \]
Thus, we have:
\[ -\ln(2) = -\lambda T_{1/2} \]

### Step 7: Solve for Half-Life
Now, we can solve for \( T_{1/2} \):
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]

### Conclusion
The relation between half-life period \( T_{1/2} \) and disintegration constant \( \lambda \) is given by:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]