Atomic weight of boron is `10.81` and it has two isotopes `._5 B^10` and `._5 B^11`. Then ratio of `._5 B^10` in nature would be.

Options

A. `19 : 81`
B. `10 : 11`
C. `15 : 16`
D. `81 : 19`

Answer

To find the ratio of the isotopes \( _5B^{10} \) and \( _5B^{11} \) in nature, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = number of \( _5B^{10} \) atoms
- \( y \) = number of \( _5B^{11} \) atoms

### Step 2: Write the Equation for Atomic Weight
The average atomic weight of boron is given as 10.81. The contribution to the atomic weight from each isotope can be expressed as:
\[ \text{Total weight} = \frac{10x + 11y}{x + y} \]
This equation represents the average atomic weight based on the number of each isotope.

### Step 3: Set Up the Equation
Setting the average atomic weight equal to the given atomic weight:
\[ \frac{10x + 11y}{x + y} = 10.81 \]

### Step 4: Cross Multiply to Eliminate the Fraction
Cross multiplying gives:
\[ 10x + 11y = 10.81(x + y) \]

### Step 5: Expand and Rearrange the Equation
Expanding the right side:
\[ 10x + 11y = 10.81x + 10.81y \]
Rearranging the equation:
\[ 10x + 11y - 10.81x - 10.81y = 0 \]
This simplifies to:
\[ -0.81x + 0.19y = 0 \]

### Step 6: Solve for the Ratio
Rearranging gives:
\[ 0.81x = 0.19y \]
Dividing both sides by \( y \) and \( x \):
\[ \frac{x}{y} = \frac{0.19}{0.81} \]
To simplify this ratio, we can multiply both the numerator and the denominator by 100:
\[ \frac{x}{y} = \frac{19}{81} \]

### Step 7: Conclusion
Thus, the ratio of \( _5B^{10} \) to \( _5B^{11} \) in nature is:
\[ \text{Ratio of } _5B^{10} : _5B^{11} = 19 : 81 \]