A radioactive substance has `10^8` nuclei. Its half life is 30 s . The number of nuclei left after 15 s is nearly

Options

A. `2 xx 10^5`
B. `3 xx 10^6`
C. `7xx 10^7`
D. `5 xx 10^8`

Answer

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

### Step 1: Understand the given data
- Initial number of nuclei, \( N_0 = 10^8 \)
- Half-life, \( T_{1/2} = 30 \) seconds
- Time elapsed, \( t = 15 \) seconds

### Step 2: Calculate the decay constant (\( \lambda \))
The decay constant (\( \lambda \)) can be calculated using the formula:
\[ \lambda = \frac{\ln 2}{T_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{\ln 2}{30 \text{ s}} \]

### Step 3: Use the radioactive decay formula
The number of nuclei remaining after time \( t \) can be calculated using the formula:
\[ N_t = N_0 e^{-\lambda t} \]
Substituting the values:
\[ N_t = 10^8 e^{-\lambda \cdot 15} \]

### Step 4: Substitute \( \lambda \) into the equation
Now, substituting \( \lambda \):
\[ N_t = 10^8 e^{-\left(\frac{\ln 2}{30}\right) \cdot 15} \]
This simplifies to:
\[ N_t = 10^8 e^{-\frac{15 \ln 2}{30}} = 10^8 e^{-\frac{1}{2} \ln 2} \]

### Step 5: Simplify the expression
Using the property of exponents:
\[ e^{-\frac{1}{2} \ln 2} = 2^{-\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ N_t = 10^8 \cdot \frac{1}{\sqrt{2}} = \frac{10^8}{\sqrt{2}} \]

### Step 6: Calculate the numerical value
Calculating \( \frac{10^8}{\sqrt{2}} \):
\[ \sqrt{2} \approx 1.414 \]
So,
\[ N_t \approx \frac{10^8}{1.414} \approx 7.07 \times 10^7 \]

### Step 7: Conclusion
The number of nuclei left after 15 seconds is approximately \( 7.07 \times 10^7 \). The nearest option to this value is \( 7 \times 10^7 \).

### Final Answer
Thus, the number of nuclei left after 15 seconds is nearly \( 7 \times 10^7 \).

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