Question

An element of avg. atomic mass z, has a isotopic masses

(z + 2) and (z – 1). Therefore percentage abundance of the heavier isotope is –

• A. $\frac{100}{3}$%
• B. $\frac{100}{2}$%
• C. $\frac{100}{4}$%
• D. $\frac{(z + 1)}{(z–1)}$× 100%

Answer 1

Answer: (A)

$\frac{100}{3}$%

Answer 2

Let % abundance of heavier isotope is x %

$\therefore$ $\frac{(z + 2) \times x + (z–1)(100–x)}{100}$ = z

or xz + 2x – xz + x + 100z – 100 = 100z

or 3x = 100 or x = $\frac{100}{3}$%