The probability that particular nucleus of `""_(38)Cl` will undergo `beta`-decay in time interval `10^(-3)` sec (the half life of `""_(38)Cl` is 37.2 min) is

Options

A. `3.1 xx 10^(-7)`
B. `2.25 xx 10^(-6)`
C. `3.36 xx 10^(-5)`
D. `1.73 xx 10^(-7)`

Answer

To find the probability that a particular nucleus of \( _{38}^{Cl} \) will undergo beta decay in a time interval of \( 10^{-3} \) seconds, we can follow these steps:

### Step 1: Understand the relationship between half-life and decay constant
The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given that the half-life of \( _{38}^{Cl} \) is 37.2 minutes, we first convert this into seconds:
\[ t_{1/2} = 37.2 \, \text{minutes} \times 60 \, \text{seconds/minute} = 2232 \, \text{seconds} \]

### Step 2: Calculate the decay constant \( \lambda \)
Now we can calculate \( \lambda \):
\[ \lambda = \frac{\ln(2)}{2232} \approx \frac{0.693}{2232} \approx 0.0003105 \, \text{s}^{-1} \]

### Step 3: Use the decay formula to find the number of undecayed nuclei
The number of undecayed nuclei \( N(t) \) at time \( t \) is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei. For a time interval \( t = 10^{-3} \) seconds, we have:
\[ N(10^{-3}) = N_0 e^{-\lambda \cdot 10^{-3}} = N_0 e^{-0.0003105 \cdot 10^{-3}} \]

### Step 4: Calculate the probability of decay
The probability \( P \) that a nucleus will decay in the time interval \( t \) is given by:
\[ P = 1 - \frac{N(t)}{N_0} = 1 - e^{-\lambda t} \]
Substituting the values we have:
\[ P = 1 - e^{-0.0003105 \cdot 10^{-3}} \]

### Step 5: Calculate the numerical value of the probability
Now we can calculate \( P \):
\[ P \approx 1 - e^{-0.0000003105} \approx 1 - (1 - 0.0000003105) \approx 0.0000003105 \]
Thus,
\[ P \approx 3.1 \times 10^{-7} \]

### Final Answer
The probability that a particular nucleus of \( _{38}^{Cl} \) will undergo beta decay in a time interval of \( 10^{-3} \) seconds is approximately:
\[ P \approx 3.1 \times 10^{-7} \]