In nature, ratio of isotopes Boron, `""_5B^(10) and ""_5B^(11)` , is (given that atomic weight of boron is 10.81)

Options

A. `81:19`
B. `21:44`
C. `19:81`
D. `44:21`

Answer

To find the ratio of the isotopes \( _5B^{10} \) and \( _5B^{11} \) in nature, we can use the information given about the atomic weight of boron, which is 10.81.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the fraction (in percentage) of the isotope \( _5B^{10} \) in nature. Therefore, the fraction of the isotope \( _5B^{11} \) will be \( 100 - x \).

2. **Set Up the Equation**:
The average atomic weight of boron can be expressed as:
\[ \text{Atomic weight} = \frac{x \cdot 10 + (100 - x) \cdot 11}{100} \]
Given that the atomic weight of boron is 10.81, we can set up the equation:
\[ \frac{x \cdot 10 + (100 - x) \cdot 11}{100} = 10.81 \]

3. **Multiply Through by 100**:
To eliminate the fraction, multiply both sides by 100:
\[ x \cdot 10 + (100 - x) \cdot 11 = 1081 \]

4. **Expand and Simplify**:
Expanding the left side gives:
\[ 10x + 1100 - 11x = 1081 \]
Simplifying this results in:
\[ -x + 1100 = 1081 \]

5. **Solve for \( x \)**:
Rearranging gives:
\[ -x = 1081 - 1100 \]
\[ -x = -19 \quad \Rightarrow \quad x = 19 \]

6. **Find the Percentage of Each Isotope**:
Since \( x = 19 \), the percentage of \( _5B^{10} \) is 19%, and the percentage of \( _5B^{11} \) is:
\[ 100 - x = 100 - 19 = 81\% \]

7. **Calculate the Ratio**:
The ratio of the isotopes \( _5B^{10} \) to \( _5B^{11} \) is:
\[ \text{Ratio} = \frac{19}{81} \]

### Final Answer:
The ratio of isotopes \( _5B^{10} \) to \( _5B^{11} \) in nature is \( 19:81 \).

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