1 gram of a radioactive element reduces to `1/3` gram at the end of 2 days. Then the mass of the element remaining at the end of 6 days is- [gram]

Options

A. `1/12`
B. `1/27`
C. `1/6`
D. `1/9`

Answer

To solve the problem, we need to determine the mass of the radioactive element remaining after 6 days, given that 1 gram reduces to 1/3 gram in 2 days.

### Step-by-Step Solution:

1. **Understanding the Decay Formula**:
The decay of a radioactive substance can be described by the formula:
\[ N = N_0 e^{-\lambda t} \]
where:
- \( N \) is the remaining quantity of the substance,
- \( N_0 \) is the initial quantity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

2. **Setting Up the Initial Condition**:
We know that:
- Initial mass \( N_0 = 1 \) gram,
- Mass after 2 days \( N = \frac{1}{3} \) gram,
- Time \( t = 2 \) days.

Plugging these values into the decay formula gives:
\[ \frac{1}{3} = 1 \cdot e^{-\lambda \cdot 2} \]

3. **Solving for the Decay Constant \( \lambda \)**:
Rearranging the equation:
\[ e^{-2\lambda} = \frac{1}{3} \]
Taking the natural logarithm on both sides:
\[ -2\lambda = \ln\left(\frac{1}{3}\right) \]
Thus,
\[ \lambda = -\frac{1}{2} \ln\left(\frac{1}{3}\right) = \frac{\ln(3)}{2} \]

4. **Finding the Remaining Mass After 6 Days**:
Now, we want to find the mass remaining after 6 days. Using the decay formula again:
\[ N = N_0 e^{-\lambda t} \]
Here, \( t = 6 \) days:
\[ N = 1 \cdot e^{-\left(\frac{\ln(3)}{2}\right) \cdot 6} \]
Simplifying:
\[ N = e^{-3 \ln(3)} = e^{\ln(3^{-3})} = 3^{-3} = \frac{1}{27} \]

5. **Final Result**:
Therefore, the mass of the radioactive element remaining at the end of 6 days is:
\[ N = \frac{1}{27} \text{ grams} \]

### Summary:
The mass of the radioactive element remaining at the end of 6 days is \( \frac{1}{27} \) grams.