Question
Question: Boron has two isotopes, B-10 and B-11, the average atomic mass of boron is found to be \( {\text{10}...
Boron has two isotopes, B-10 and B-11, the average atomic mass of boron is found to be 1080mu . Calculate the percentage of abundance of these isotopes.
Solution
In the above question, boron’s two isotopes are given along with the average atomic mass. We have to find the percentage of these isotopes. We can put the value of two isotopes' atomic mass in the average atomic mass equation to get the percentage of two isotopes.
Formula Used
A = i = 1∑naipi
Where A = atomic mass of the element
ai =atomic mass of ith isotope.
pi = percentage of occurrence of ith isotope.
n= number of isotopes.
Complete step by step solution:
Isotope of a particular element can be defined as the element having the same atomic number but different atomic mass. In other words, we can say that isotopes have the same number of protons but they differ in the number of neutrons.
Boron has two isotopes B-10 and B-11.
Atomic mass of B-10 is 10 and atomic mass of B-11 is 11.Let the percentage of B-10 be x and the percentage of B-11 be 1−x .
The average atomic mass can be written as:
A = i = 1∑naipi
Since, boron has only 2 isotopes and hence we can replace n with 2.
A = i = 1∑2aipi = a1p1 + a2p2
Substituting the values, we get:
10.8=10×x+11×(1−x)
⇒10.8 = 10x + 11−11x
Rearranging the equation, we get:
x = 11−10.8 = 0.2
Now we can find the value of 1−x :
1−x = 1−0.2 = 0.8
Therefore, the percentage of abundances of B-10 and B-11 is {\text{20% }} and {\text{80% }} respectively.
Note:
In these types of questions, where we have to calculate the percentage abundance of the isotopes of an element we have to take their atomic mass of their isotopes. In case, we have only 2 isotopes, we can take the percentage of one isotope as x and other as 1−x .