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 of Boron, \( \text{B}^{10} \) and \( \text{B}^{11} \), given that the atomic weight of boron is 10.81, we can follow these steps:

### Step 1: Define the variables
Let:
- \( X_1 \) = percentage of \( \text{B}^{10} \)
- \( X_2 \) = percentage of \( \text{B}^{11} \)

Since these are the only two isotopes of boron, we have:
\[ X_1 + X_2 = 100 \]

### Step 2: Set up the equation for atomic weight
The average atomic weight of boron can be expressed as:
\[ \text{Atomic Weight} = \frac{(A_1 \cdot X_1) + (A_2 \cdot X_2)}{100} \]
Where:
- \( A_1 = 10 \) (atomic mass of \( \text{B}^{10} \))
- \( A_2 = 11 \) (atomic mass of \( \text{B}^{11} \))

Substituting the values:
\[ 10.81 = \frac{(10 \cdot X_1) + (11 \cdot X_2)}{100} \]

### Step 3: Rearranging the equation
Multiply both sides by 100:
\[ 10.81 \cdot 100 = 10X_1 + 11X_2 \]
\[ 1081 = 10X_1 + 11X_2 \] (Equation 1)

### Step 4: Substitute \( X_1 \) in terms of \( X_2 \)
From the first equation, we can express \( X_1 \):
\[ X_1 = 100 - X_2 \] (Equation 2)

### Step 5: Substitute Equation 2 into Equation 1
Substituting \( X_1 \) in Equation 1:
\[ 1081 = 10(100 - X_2) + 11X_2 \]
\[ 1081 = 1000 - 10X_2 + 11X_2 \]
\[ 1081 = 1000 + X_2 \]

### Step 6: Solve for \( X_2 \)
Rearranging gives:
\[ X_2 = 1081 - 1000 \]
\[ X_2 = 81 \]

### Step 7: Solve for \( X_1 \)
Now substitute \( X_2 \) back into Equation 2:
\[ X_1 = 100 - 81 \]
\[ X_1 = 19 \]

### Step 8: Find the ratio \( X_1 : X_2 \)
Now, we can find the ratio of \( X_1 \) to \( X_2 \):
\[ \text{Ratio} = \frac{X_1}{X_2} = \frac{19}{81} \]

Thus, the ratio of the isotopes \( \text{B}^{10} \) to \( \text{B}^{11} \) is:
\[ 19 : 81 \]

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

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