Vanadium has two naturally occurring isotopes, 50V with an atomic mass of 49.9472 amu and 51y with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are %51V % 50V and %51V
The Correct Answer and Explanation is :
To solve this problem, let ( x ) represent the percent abundance of ( ^{50}\text{V} ), and ( 1 – x ) represent the percent abundance of ( ^{51}\text{V} ). The atomic weight of vanadium is the weighted average of the atomic masses of its isotopes:
[
\text{Atomic weight} = (x \times 49.9472) + [(1 – x) \times 50.9440]
]
The atomic weight is given as 50.9415, so the equation becomes:
[
50.9415 = (x \times 49.9472) + [(1 – x) \times 50.9440]
]
Step 1: Simplify the equation
Expand the terms:
[
50.9415 = (x \times 49.9472) + (50.9440 – x \times 50.9440)
]
[
50.9415 = (49.9472x) + 50.9440 – (50.9440x)
]
Combine like terms:
[
50.9415 = 50.9440 – 0.9968x
]
Step 2: Solve for ( x )
Rearrange to isolate ( x ):
[
0.9968x = 50.9440 – 50.9415
]
[
0.9968x = 0.0025
]
[
x = \frac{0.0025}{0.9968}
]
[
x \approx 0.00251
]
So, the percent abundance of ( ^{50}\text{V} ) is approximately:
[
x \times 100 \approx 0.251\%
]
The percent abundance of ( ^{51}\text{V} ) is:
[
(1 – x) \times 100 \approx 99.749\%
]
Final Answer:
- Percent abundance of ( ^{50}\text{V} ): 0.251%
- Percent abundance of ( ^{51}\text{V} ): 99.749%
Explanation:
The atomic weight of an element is a weighted average of its isotopic masses, determined by their relative abundances. For vanadium, ( ^{50}\text{V} ) is much less abundant than ( ^{51}\text{V} ), as indicated by their atomic weights and the overall atomic weight. Solving the equation shows ( ^{51}\text{V} ) is almost 100% of natural vanadium, which aligns with experimental data for the element’s isotopic composition.