The half-life of a particular radioactive isotope is 6.5 h. If there are initially `65 xx 10^(19)` atoms of this isotope, how many remain at the end of 26 h?

Answer

To solve the problem of how many radioactive atoms remain after 26 hours given the half-life of the isotope is 6.5 hours and the initial number of atoms is \(65 \times 10^{19}\), we can follow these steps:

### Step 1: Determine the number of half-lives that have passed
The total time given is 26 hours, and the half-life of the isotope is 6.5 hours. We can find the number of half-lives by dividing the total time by the half-life.

\[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{26 \text{ hours}}{6.5 \text{ hours}} = 4 \]

### Step 2: Calculate the remaining number of atoms
The number of atoms remaining after a certain number of half-lives can be calculated using the formula:

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

where:
- \(N_t\) = number of remaining atoms
- \(N_0\) = initial number of atoms
- \(n\) = number of half-lives

Substituting the known values:

\[ N_t = 65 \times 10^{19} \left(\frac{1}{2}\right)^4 \]

### Step 3: Simplify the expression
Calculating \(\left(\frac{1}{2}\right)^4\):

\[ \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

Now substituting this back into the equation:

\[ N_t = 65 \times 10^{19} \times \frac{1}{16} \]

### Step 4: Perform the division
Now we can divide \(65\) by \(16\):

\[ N_t = \frac{65}{16} \times 10^{19} \]

Calculating \(\frac{65}{16}\):

\[ \frac{65}{16} = 4.0625 \]

Thus, we have:

\[ N_t = 4.0625 \times 10^{19} \]

### Final Answer
Therefore, the number of radioactive atoms remaining after 26 hours is approximately:

\[ N_t \approx 4.06 \times 10^{19} \text{ atoms} \]

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