
Vanadium has two naturally occurring isotopes, $ ^{50}V $ with an atomic mass of 49.9472 amu and $ ^{51}V $ with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. What is the percent abundance of the vanadium isotopes?
Answer
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Hint: Isotopes are the species that have the same atomic number but have different atomic mass. Isotopes do not have separate spaces in the periodic table as the atomic number is the same. Isotopes are formed because of the difference in the no. of neutrons in the atom of an element.
Complete answer:
We are given that Vanadium has two isotopes $ ^{50}V $ and $ ^{51}V $ . This means that the total isotopic abundance of these two isotopes will be 100% . Here we’ll use decimal abundances which are the percentage abundances divided by 100 for easy calculations.
Since we are taking decimal abundances the, the decimal abundance of one isotope be ‘x’ and that of the another isotope will automatically be equal to (1-x), since the total of the two abundances will be: $ x + (1 - x) = 1 $
The two decimal abundances will add up to give one. An element’s relative atomic mass is calculated by taking the average of the atomic masses of its naturally occurring isotope. Relative atomic mass can hence be given as:
$ {\text{relative atomic mass}} = \sum\limits_i {(isotop{e_i} \times abundance{e_i})} $
So let us consider the decimal abundance of $ ^{50}V $ to be x and that of $ ^{51}V $ to be (1-x). The atomic masses of the isotopes are given to us and the atomic mass of the element Vanadium is 50.9415. According to the formula above the relative atomic mass can be given as:
$ 50.9415 = (49.9472 \times x) + 50.9440(1 - x) $
$ 50.9415 = 49.9472x + 50.9440 - 50.9440x $
$ 0.0025 = 0.9968x \Rightarrow x = \dfrac{{0.0025}}{{0.9968}} = 0.00250 $
X is the decimal abundance of $ ^{50}V $ , therefore the decimal abundance of $ ^{51}V $ will be $ = 1 - 0.00250 = 0.99749 $
To find the percentage abundance we will multiply the values by 100.
Therefore, the percentage abundance of $ ^{50}V $ $ = 0.00250 \times 100 = 0.250\% $ and the percentage abundance of $ ^{51}V $ $ = 0.99749 \times 100 = 99.749\% $
This is the required answer.
Note:
Isotopes differ in the number of neutrons, the number of protons and electrons remain the same. One element can have more than 2 isotopes also. Example Hydrogen has three isotopes namely, Deuterium and tritium denoted as $ _1^2H\& _1^3H $ .
Complete answer:
We are given that Vanadium has two isotopes $ ^{50}V $ and $ ^{51}V $ . This means that the total isotopic abundance of these two isotopes will be 100% . Here we’ll use decimal abundances which are the percentage abundances divided by 100 for easy calculations.
Since we are taking decimal abundances the, the decimal abundance of one isotope be ‘x’ and that of the another isotope will automatically be equal to (1-x), since the total of the two abundances will be: $ x + (1 - x) = 1 $
The two decimal abundances will add up to give one. An element’s relative atomic mass is calculated by taking the average of the atomic masses of its naturally occurring isotope. Relative atomic mass can hence be given as:
$ {\text{relative atomic mass}} = \sum\limits_i {(isotop{e_i} \times abundance{e_i})} $
So let us consider the decimal abundance of $ ^{50}V $ to be x and that of $ ^{51}V $ to be (1-x). The atomic masses of the isotopes are given to us and the atomic mass of the element Vanadium is 50.9415. According to the formula above the relative atomic mass can be given as:
$ 50.9415 = (49.9472 \times x) + 50.9440(1 - x) $
$ 50.9415 = 49.9472x + 50.9440 - 50.9440x $
$ 0.0025 = 0.9968x \Rightarrow x = \dfrac{{0.0025}}{{0.9968}} = 0.00250 $
X is the decimal abundance of $ ^{50}V $ , therefore the decimal abundance of $ ^{51}V $ will be $ = 1 - 0.00250 = 0.99749 $
To find the percentage abundance we will multiply the values by 100.
Therefore, the percentage abundance of $ ^{50}V $ $ = 0.00250 \times 100 = 0.250\% $ and the percentage abundance of $ ^{51}V $ $ = 0.99749 \times 100 = 99.749\% $
This is the required answer.
Note:
Isotopes differ in the number of neutrons, the number of protons and electrons remain the same. One element can have more than 2 isotopes also. Example Hydrogen has three isotopes namely, Deuterium and tritium denoted as $ _1^2H\& _1^3H $ .
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