The natural boron of atomic weight 10.81 is found to have two isotopes `.^10B` and `.^11B` .The ratio of abundance of isotopes of natural boron should be

Options

A. `11:10`
B. `81:19`
C. `10:11`
D. `19:81`

Explanation

To find the ratio of abundance of the isotopes of natural boron, we can follow these steps:

### Step 1: Define Variables
Let the abundance of the isotope \( ^{10}B \) be \( X \% \). Consequently, the abundance of the isotope \( ^{11}B \) will be \( (100 - X) \% \).

### Step 2: Set Up the Equation
The average atomic weight of boron is given as 10.81. We can express this in terms of the abundances and atomic weights of the isotopes:
\[ 10.81 = \left( \frac{10 \cdot X + 11 \cdot (100 - X)}{100} \right) \]

### Step 3: Simplify the Equation
Multiply both sides by 100 to eliminate the fraction:
\[ 10.81 \cdot 100 = 10X + 11(100 - X) \]
\[ 1081 = 10X + 1100 - 11X \]

### Step 4: Combine Like Terms
Rearranging the equation gives:
\[ 1081 = 1100 - X \]
\[ X = 1100 - 1081 \]
\[ X = 19 \]

### Step 5: Calculate the Abundance of \( ^{11}B \)
Now we can find the abundance of \( ^{11}B \):
\[ \text{Abundance of } ^{11}B = 100 - X = 100 - 19 = 81 \]

### Step 6: Find the Ratio
Now we can find the ratio of the abundances of \( ^{10}B \) to \( ^{11}B \):
\[ \text{Ratio} = \frac{X}{100 - X} = \frac{19}{81} \]

### Final Answer
The ratio of the abundance of isotopes \( ^{10}B \) to \( ^{11}B \) is \( \frac{19}{81} \).

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