The half life of the radioactive element is 35 years. If there are `4xx10^(6)` nuclei at the start then after how many years `0.5xx10^(6)` of them would be left ?

Options

A. 35
B. 70
C. 105
D. 140

Explanation

To solve the problem step by step, we will use the concept of half-life in radioactive decay.

### Step 1: Understand the given data
- The half-life (t₁/₂) of the radioactive element is 35 years.
- The initial number of nuclei (N₀) is \(4 \times 10^6\).
- We need to find out after how many years (t) there will be \(0.5 \times 10^6\) nuclei left (N).

### Step 2: Use the half-life formula
The relationship between the initial quantity (N₀), the remaining quantity (N), and the number of half-lives (n) is given by the formula:

\[ N = N_0 \left(\frac{1}{2}\right)^n \]

### Step 3: Set up the equation
We know:
- \(N = 0.5 \times 10^6\)
- \(N_0 = 4 \times 10^6\)

Substituting these values into the equation gives:

\[ 0.5 \times 10^6 = 4 \times 10^6 \left(\frac{1}{2}\right)^n \]

### Step 4: Simplify the equation
Dividing both sides by \(4 \times 10^6\):

\[ \frac{0.5 \times 10^6}{4 \times 10^6} = \left(\frac{1}{2}\right)^n \]

This simplifies to:

\[ \frac{0.5}{4} = \left(\frac{1}{2}\right)^n \]

\[ \frac{1}{8} = \left(\frac{1}{2}\right)^n \]

### Step 5: Relate to powers of 2
We know that:

\[ \frac{1}{8} = \left(\frac{1}{2}\right)^3 \]

Thus, we can equate the exponents:

\[ n = 3 \]

### Step 6: Calculate the time
Since each half-life is 35 years, the total time (t) taken for 3 half-lives is:

\[ t = n \times t_{1/2} = 3 \times 35 \text{ years} = 105 \text{ years} \]

### Final Answer
After 105 years, \(0.5 \times 10^6\) of the nuclei would be left.

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