There are two isotopes of an element with atomic mass z. Heavier on has atomic mass z+2 and lighter one has z-1, then abundance of lighter one is

Options

A. `66.6%`
B. `96.7%`
C. `6.67%`
D. `33.3%`

Explanation

To find the abundance of the lighter isotope, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = percent abundance of the lighter isotope
- \( 100 - x \) = percent abundance of the heavier isotope

### Step 2: Write the Average Atomic Mass Equation
The average atomic mass of the isotopes can be expressed as:
\[ \text{Average atomic mass} = \frac{(x \cdot (z - 1)) + ((100 - x) \cdot (z + 2))}{100} \]
Given that the average atomic mass is \( z \), we can set up the equation:
\[ \frac{(x \cdot (z - 1)) + ((100 - x) \cdot (z + 2))}{100} = z \]

### Step 3: Multiply Both Sides by 100
To eliminate the denominator, multiply both sides by 100:
\[ (x \cdot (z - 1)) + ((100 - x) \cdot (z + 2)) = 100z \]

### Step 4: Expand the Equation
Expanding the left side gives:
\[ x(z - 1) + (100 - x)(z + 2) = 100z \]
\[ xz - x + 100z + 200 - xz - 2x = 100z \]

### Step 5: Simplify the Equation
Combine like terms:
\[ 100z + 200 - 3x = 100z \]
Subtract \( 100z \) from both sides:
\[ 200 - 3x = 0 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ 3x = 200 \]
\[ x = \frac{200}{3} \approx 66.67 \]

### Conclusion
The abundance of the lighter isotope is approximately \( 66.67\% \).