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 find the value of \( n \) given the half-life of a radioactive element \( A \) and the amount left after a certain period. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Data
- Half-life (\( t_{1/2} \)) of element \( A \) = 10 days
- Initial amount of \( A \) (\( A_0 \)) = 1 mole
- Time elapsed (\( t \)) = 11 days
- Amount left after 11 days = \( \left(\frac{1}{2}\right)^{\frac{n+6}{10}} \) moles

### Step 2: Calculate the Decay Constant (\( k \))
The decay constant \( k \) can be calculated using the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the value of \( t_{1/2} \):
\[ k = \frac{0.693}{10} = 0.0693 \, \text{day}^{-1} \]

### Step 3: Use the Radioactive Decay Formula
The amount of substance remaining after time \( t \) is given by:
\[ A = A_0 e^{-kt} \]
However, we can also express it in logarithmic form:
\[ \frac{A_0}{A} = 2^{\frac{t}{t_{1/2}}} \]
Substituting the values:
\[ \frac{1}{A} = 2^{\frac{11}{10}} = 2^{1.1} \]

### Step 4: Set Up the Equation
From the problem statement, we have:
\[ A = \left(\frac{1}{2}\right)^{\frac{n+6}{10}} \]
Thus, we can write:
\[ \frac{1}{\left(\frac{1}{2}\right)^{\frac{n+6}{10}}} = 2^{\frac{11}{10}} \]
This simplifies to:
\[ 2^{\frac{n+6}{10}} = 2^{1.1} \]

### Step 5: Equate the Exponents
Since the bases are the same, we can equate the exponents:
\[ \frac{n+6}{10} = 1.1 \]

### Step 6: Solve for \( n \)
To find \( n \), multiply both sides by 10:
\[ n + 6 = 11 \]
Now, subtract 6 from both sides:
\[ n = 5 \]

### Final Answer
The value of \( n \) is \( 5 \).

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