Half - life period of the radioactive element A is 10 days. Amount of A left on the end of 11th day staring with 1 mole A is `((1)/(2))^((n+6)/(10))` mole. What is the value of n.

Answer

To solve the problem, we need to determine the value of \( n \) based on the information provided about the radioactive element A. The half-life of A is given as 10 days, and we need to find out how much of A is left after 11 days starting with 1 mole.

### Step-by-step Solution:

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

2. **Determine the Decay Constant (k)**:
The decay constant \( k \) can be calculated using the formula:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{10} = 0.0693 \, \text{days}^{-1} \]

3. **Calculate the Remaining Amount After 11 Days**:
The formula for the amount of radioactive substance remaining after time \( t \) is given by:
\[ N_t = N_0 e^{-kt} \]
where \( N_0 \) is the initial amount, \( N_t \) is the amount remaining after time \( t \), and \( k \) is the decay constant.

Since we are starting with 1 mole of A:
\[ N_t = 1 \cdot e^{-0.0693 \times 11} \]

4. **Substituting Values**:
Calculate \( e^{-0.0693 \times 11} \):
\[ e^{-0.0693 \times 11} = e^{-0.7623} \approx 0.466 \]
Therefore, the amount of A left after 11 days is approximately 0.466 moles.

5. **Relate to the Given Expression**:
The problem states that the amount left can also be expressed as:
\[ N_t = \left(\frac{1}{2}\right)^{\frac{n + 6}{10}} \]
Setting the two expressions for \( N_t \) equal gives:
\[ 0.466 = \left(\frac{1}{2}\right)^{\frac{n + 6}{10}} \]

6. **Convert to Logarithmic Form**:
To solve for \( n \), we can take the logarithm:
\[ \log(0.466) = \frac{n + 6}{10} \log\left(\frac{1}{2}\right) \]

7. **Calculate Logarithms**:
We know that \( \log\left(\frac{1}{2}\right) = -\log(2) \). Using \( \log(2) \approx 0.301 \):
\[ \log(0.466) \approx -0.332 \]
Thus, we have:
\[ -0.332 = \frac{n + 6}{10} \times -0.301 \]

8. **Solve for n**:
Rearranging gives:
\[ 0.332 = \frac{n + 6}{10} \times 0.301 \]
Multiplying both sides by 10:
\[ 3.32 = (n + 6) \times 0.301 \]
Dividing by 0.301:
\[ n + 6 \approx \frac{3.32}{0.301} \approx 11.02 \]
Therefore:
\[ n \approx 11.02 - 6 = 5.02 \]

9. **Final Answer**:
Rounding gives \( n \approx 5 \).

### Conclusion:
The value of \( n \) is **5**.

---