The half life a radioactive element is `30` days, if the mass of the specimen reduces to `1//10^(th)` then the time taken is :-

Options

A. `100` days
B. `50` days
C. `33` days
D. `16` days

Explanation

To solve the problem, we need to determine the time taken for the mass of a radioactive specimen to reduce to \( \frac{1}{10} \) of its original mass, given that its half-life is 30 days.

### Step-by-Step Solution:

1. **Understanding Half-Life**:
The half-life of a radioactive element is the time taken for half of the radioactive atoms in a sample to decay. In this case, the half-life \( t_{1/2} \) is given as 30 days.

2. **Decay Constant Calculation**:
The decay constant \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{\ln(2)}{30 \text{ days}} \]

3. **Using the Exponential Decay Formula**:
The number of radioactive atoms remaining after time \( t \) can be expressed as:
\[ N(t) = N_0 e^{-\lambda t} \]
Where \( N_0 \) is the initial quantity of the substance.

4. **Setting Up the Equation**:
We want to find the time \( t \) when the mass reduces to \( \frac{1}{10} \) of its original mass:
\[ N(t) = \frac{N_0}{10} \]
Substituting into the decay formula:
\[ \frac{N_0}{10} = N_0 e^{-\lambda t} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{10} = e^{-\lambda t} \]

5. **Taking Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{10}\right) = -\lambda t \]
This can be rewritten as:
\[ -\ln(10) = -\lambda t \]
Thus:
\[ t = \frac{\ln(10)}{\lambda} \]

6. **Substituting the Decay Constant**:
Substitute \( \lambda \) into the equation:
\[ t = \frac{\ln(10)}{\frac{\ln(2)}{30}} = 30 \cdot \frac{\ln(10)}{\ln(2)} \]

7. **Calculating Values**:
Using approximate values:
- \( \ln(10) \approx 2.303 \)
- \( \ln(2) \approx 0.693 \)

Now substituting these values:
\[ t \approx 30 \cdot \frac{2.303}{0.693} \approx 30 \cdot 3.32 \approx 99.6 \text{ days} \]

8. **Final Result**:
Rounding off, we find that the time taken for the mass to reduce to \( \frac{1}{10} \) of its original mass is approximately 100 days.

### Conclusion:
The time taken for the mass of the specimen to reduce to \( \frac{1}{10} \) of its original mass is **100 days**.