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`

Explanation

To find the ratio of the isotopes \( _5B^{10} \) and \( _5B^{11} \) in nature, we can use the concept of average atomic weight and the contributions of each isotope to that average. Here’s a step-by-step solution:

### Step 1: Define the variables
Let:
- \( x \) = percentage of \( _5B^{10} \)
- \( 100 - x \) = percentage of \( _5B^{11} \)

### Step 2: Write the equation for average atomic weight
The average atomic weight of boron is given as 10.81. The average atomic weight can be calculated using the contributions from each isotope:
\[ \text{Average Atomic Weight} = \frac{(10 \cdot x) + (11 \cdot (100 - x))}{100} \]
Setting this equal to 10.81 gives us:
\[ \frac{(10x) + (11(100 - x))}{100} = 10.81 \]

### Step 3: Simplify the equation
Multiply both sides by 100 to eliminate the denominator:
\[ 10x + 1100 - 11x = 1081 \]
Combine like terms:
\[ -1x + 1100 = 1081 \]

### Step 4: Solve for \( x \)
Rearranging gives:
\[ -1x = 1081 - 1100 \]
\[ -1x = -19 \]
\[ x = 19 \]

### Step 5: Calculate the percentage of \( _5B^{11} \)
Since \( x = 19 \), the percentage of \( _5B^{11} \) is:
\[ 100 - x = 100 - 19 = 81 \]

### Step 6: Find the ratio of the isotopes
The ratio of \( _5B^{10} \) to \( _5B^{11} \) is:
\[ \text{Ratio} = \frac{x}{100 - x} = \frac{19}{81} \]

### Conclusion
Thus, the ratio of \( _5B^{10} \) to \( _5B^{11} \) in nature is \( 19:81 \).

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