Half-life of a radioacitve element is `10` days. The time during which quantity remains `1//10` of initial mass will be

Options

A. 99 days
B. 33 days
C. 16 days
D. 70 days

Answer

To solve the problem, we need to find the time during which the quantity of a radioactive element remains \( \frac{1}{10} \) of its initial mass, given that its half-life is \( 10 \) days.

### Step-by-Step Solution:

1. **Understanding Half-Life**:
The half-life (\( t_{1/2} \)) of a radioactive element is the time required for half of the radioactive substance to decay. Here, \( t_{1/2} = 10 \) days.

2. **Initial Mass**:
Let's denote the initial mass of the radioactive element as \( N_0 \).

3. **Final Mass**:
We want to find the time \( t \) when the remaining mass is \( \frac{N_0}{10} \).

4. **Using the Decay Formula**:
The amount of substance remaining after time \( t \) can be expressed using the formula:
\[ N(t) = N_0 \cdot e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

5. **Relating Half-Life to Decay Constant**:
The decay constant \( \lambda \) can be calculated from the half-life using the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Substituting \( t_{1/2} = 10 \) days:
\[ \lambda = \frac{0.693}{10} = 0.0693 \text{ days}^{-1} \]

6. **Setting Up the Equation**:
We set up the equation for the remaining mass:
\[ \frac{N_0}{10} = N_0 \cdot e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{10} = e^{-\lambda t} \]

7. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{10}\right) = -\lambda t \]
We know that \( \ln\left(\frac{1}{10}\right) = -\ln(10) \), so:
\[ -\ln(10) = -\lambda t \]
Thus:
\[ \ln(10) = \lambda t \]

8. **Substituting for \( \lambda \)**:
Substitute \( \lambda = 0.0693 \):
\[ t = \frac{\ln(10)}{\lambda} = \frac{2.3026}{0.0693} \]

9. **Calculating Time**:
Performing the calculation:
\[ t \approx \frac{2.3026}{0.0693} \approx 33.2 \text{ days} \]

### Final Answer:
The time during which the quantity remains \( \frac{1}{10} \) of the initial mass is approximately **33 days**.