A sample of a radioactive substance has `10^(6)` radioactive nuclei. Its half life time is 20 s How many nuclei will remain after 10 s ?

Options

A. `1xx 10^5`
B. `2 xx 10^5`
C. `7xx 10^5`
D. `11 xx 10^5`

Answer

To solve the problem of how many radioactive nuclei remain after 10 seconds, we can follow these steps:

### Step 1: Understand the given data
- Initial number of radioactive nuclei, \( N_0 = 10^6 \)
- Half-life time, \( t_{1/2} = 20 \) seconds
- Time elapsed, \( T = 10 \) seconds

### Step 2: Calculate the decay constant (\( \lambda \))
The decay constant (\( \lambda \)) can be calculated using the half-life formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Substituting the value of half-life:
\[ \lambda = \frac{\ln(2)}{20 \, \text{s}} \]

### Step 3: Use the radioactive decay formula
The number of radioactive nuclei remaining after time \( T \) is given by:
\[ N_T = N_0 e^{-\lambda T} \]
Substituting the values we have:
\[ N_T = 10^6 e^{-\lambda \cdot 10} \]

### Step 4: Substitute \( \lambda \) into the equation
Now, substituting \( \lambda \):
\[ N_T = 10^6 e^{-\left(\frac{\ln(2)}{20}\right) \cdot 10} \]
This simplifies to:
\[ N_T = 10^6 e^{-\frac{10 \ln(2)}{20}} = 10^6 e^{-\frac{\ln(2)}{2}} \]

### Step 5: Simplify the exponent
Using the property of exponents, we can rewrite:
\[ e^{-\frac{\ln(2)}{2}} = e^{\ln(2^{-1/2})} = 2^{-1/2} = \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ N_T = 10^6 \cdot \frac{1}{\sqrt{2}} = \frac{10^6}{\sqrt{2}} \]

### Step 6: Calculate the final number of nuclei
Calculating \( \frac{10^6}{\sqrt{2}} \):
\[ N_T \approx 10^6 \cdot 0.7071 \approx 707106.78 \]
Rounding to significant figures, we get:
\[ N_T \approx 7.07 \times 10^5 \]

### Step 7: Conclusion
The number of radioactive nuclei remaining after 10 seconds is approximately \( 7.07 \times 10^5 \).