The half-life of `Bi^210` is `5` days. What time is taken by `(7//8)^th` part of the sample of decay ?

Options

A. 7.5 days
B. 15 days
C. 10days
D. 20 days

Answer

To solve the problem of finding the time taken for \( \frac{7}{8} \) part of the sample of \( Bi^{210} \) to decay, we can follow these steps:

### Step 1: Understand the Decay Process
The half-life of \( Bi^{210} \) is given as 5 days. This means that in 5 days, half of the sample will decay.

### Step 2: Determine Remaining Sample After Decay
If \( \frac{7}{8} \) of the sample has decayed, then the remaining sample is:
\[ N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0 \]
where \( N_0 \) is the initial amount of the sample.

### Step 3: Use the Exponential Decay Formula
The number of nuclei remaining after time \( t \) can be expressed as:
\[ N = N_0 e^{-\lambda t} \]
Substituting \( N = \frac{1}{8}N_0 \) into the equation gives:
\[ \frac{1}{8}N_0 = N_0 e^{-\lambda t} \]

### Step 4: Simplify the Equation
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ \frac{1}{8} = e^{-\lambda t} \]

### Step 5: Solve for \( \lambda \)
The half-life \( t_{1/2} \) is related to the decay constant \( \lambda \) by:
\[ t_{1/2} = \frac{\ln 2}{\lambda} \]
From this, we can express \( \lambda \) as:
\[ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{5 \text{ days}} \]

### Step 6: Substitute \( \lambda \) Back into the Equation
Now substituting \( \lambda \) back into the equation:
\[ \frac{1}{8} = e^{-\left(\frac{\ln 2}{5}\right)t} \]

### Step 7: Take Natural Logarithm of Both Sides
Taking the natural logarithm on both sides:
\[ \ln\left(\frac{1}{8}\right) = -\left(\frac{\ln 2}{5}\right)t \]

### Step 8: Simplify the Left Side
We know that \( \frac{1}{8} = 2^{-3} \), so:
\[ \ln\left(\frac{1}{8}\right) = \ln(2^{-3}) = -3\ln(2) \]
Thus, we have:
\[ -3\ln(2) = -\left(\frac{\ln 2}{5}\right)t \]

### Step 9: Solve for \( t \)
Cancelling \( -\ln(2) \) from both sides gives:
\[ 3 = \frac{t}{5} \]
Multiplying both sides by 5 results in:
\[ t = 15 \text{ days} \]

### Final Answer
The time taken for \( \frac{7}{8} \) part of the sample to decay is **15 days**.

---