At a given instant, 60% of the radioactive nuclei in a sample are left undecayed. After 20 s, 85% nuclei have disintegrated , mean life of nuclei

Options

A. 10 s
B. 6.93 s
C. 14.43 s
D. 12.86 s

Answer

To find the mean life of the radioactive nuclei, we will follow these steps:

### Step 1: Define the initial conditions
Let the initial number of radioactive nuclei be \( N_0 \). According to the problem, at a given instant, 60% of the nuclei are undecayed. Therefore, the number of undecayed nuclei at this time is:
\[ N(t) = \frac{60}{100} N_0 = \frac{3}{5} N_0 \]

### Step 2: Determine the state after 20 seconds
After 20 seconds, it is given that 85% of the nuclei have disintegrated. This means that 15% of the nuclei remain undecayed:
\[ N(t + 20) = \frac{15}{100} N_0 = \frac{3}{20} N_0 \]

### Step 3: Use the radioactive decay formula
The number of undecayed nuclei at time \( t \) is given by the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

Using this formula for the two time points we have:
1. For \( t \):
\[ \frac{3}{5} N_0 = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ \frac{3}{5} = e^{-\lambda t} \quad \text{(1)} \]

2. For \( t + 20 \):
\[ \frac{3}{20} N_0 = N_0 e^{-\lambda (t + 20)} \]
Dividing both sides by \( N_0 \):
\[ \frac{3}{20} = e^{-\lambda (t + 20)} \quad \text{(2)} \]

### Step 4: Divide the two equations
Now, we will divide equation (1) by equation (2):
\[ \frac{\frac{3}{5}}{\frac{3}{20}} = \frac{e^{-\lambda t}}{e^{-\lambda (t + 20)}} \]
This simplifies to:
\[ \frac{3}{5} \cdot \frac{20}{3} = e^{\lambda (t + 20)} e^{-\lambda t} \]
\[ 4 = e^{\lambda (t + 20 - t)} = e^{20\lambda} \]

### Step 5: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln(4) = 20\lambda \]
We can express \( \ln(4) \) as:
\[ \ln(4) = \ln(2^2) = 2\ln(2) \]
Thus, we have:
\[ 2\ln(2) = 20\lambda \]
Solving for \( \lambda \):
\[ \lambda = \frac{2\ln(2)}{20} = \frac{\ln(2)}{10} \]

### Step 6: Calculate the mean life
The mean life \( \tau \) is related to the decay constant \( \lambda \) by:
\[ \tau = \frac{1}{\lambda} \]
Substituting the value of \( \lambda \):
\[ \tau = \frac{1}{\frac{\ln(2)}{10}} = \frac{10}{\ln(2)} \]

### Step 7: Calculate the numerical value
Using the approximate value \( \ln(2) \approx 0.693 \):
\[ \tau \approx \frac{10}{0.693} \approx 14.43 \text{ seconds} \]

### Conclusion
The mean life of the nuclei is approximately \( 14.43 \) seconds.