Answer
384.6k+ views
Hint: The relative abundance definition in science is the percentage of a specific isotope that happens in nature. The nuclear mass listed for a component on the periodic table is an average mass of all known isotopes of that component.
Complete step by step answer:
As you most likely are aware, the average nuclear mass of a component is determined by taking the weighted average of the nuclear masses of its normally occurring isotopes.
Step 1: Find the Average Atomic Mass
Basically, a component's normally occurring isotopes will contribute to the average nuclear mass of the component relative to their abundance.
\[avg.{\text{ }}atomic{\text{ }}mass\] =$\sum \left( {{\text{isotope}} \times {\text{abundance}}} \right)$
Step 2: Set Up the Relative Abundance Problem
With regards to the genuine count, it's simpler to use decimal abundances, which are basically percent abundances divided by \[100\] .
Thus, you realize that copper has two naturally occurring isotopes, \[copper - 63\] and \[copper - 65\] . This implies that their respective decimal abundance should amount to give\[1\] .
In the event that you take x to be the decimal bounty of \[copper - 63\] , you can say that the decimal abundance of \[copper - 65\] will be equivalent to \[1 - x\] .
So we can say that:
\[x \cdot 62.9296u + (1 - x) \cdot 64.9278u = 63.546u\]
Step 3: Solve for x to Get the Relative Abundance of the Unknown Isotope.
To finding the value of x we get
\[62.9296 \cdot x - 64.9278 \cdot x = 63.546 - 64.9278\] \[1.9982 \cdot x = 1.3818\]
\[x = \] $\dfrac{{1.38181}}{{0.9982}}$
\[x = {\text{ }}0.69152\]
Step 4: Find percent abundance
This implies that the percent abundances of the two isotopes will be
\[69.152\% \]---->\[^{63}Cu\]
\[30.848\% \]------.\[^{65}Cu\]
Note:
If a mass spectrum of the component was given, the relative rate isotope abundances are generally introduced as a vertical bar graph. The all-out may look as though it exceeds \[100\% ,\] however, that is because the mass spectrum works with relative rate isotope abundances.
Complete step by step answer:
As you most likely are aware, the average nuclear mass of a component is determined by taking the weighted average of the nuclear masses of its normally occurring isotopes.
Step 1: Find the Average Atomic Mass
Basically, a component's normally occurring isotopes will contribute to the average nuclear mass of the component relative to their abundance.
\[avg.{\text{ }}atomic{\text{ }}mass\] =$\sum \left( {{\text{isotope}} \times {\text{abundance}}} \right)$
Step 2: Set Up the Relative Abundance Problem
With regards to the genuine count, it's simpler to use decimal abundances, which are basically percent abundances divided by \[100\] .
Thus, you realize that copper has two naturally occurring isotopes, \[copper - 63\] and \[copper - 65\] . This implies that their respective decimal abundance should amount to give\[1\] .
In the event that you take x to be the decimal bounty of \[copper - 63\] , you can say that the decimal abundance of \[copper - 65\] will be equivalent to \[1 - x\] .
So we can say that:
\[x \cdot 62.9296u + (1 - x) \cdot 64.9278u = 63.546u\]
Step 3: Solve for x to Get the Relative Abundance of the Unknown Isotope.
To finding the value of x we get
\[62.9296 \cdot x - 64.9278 \cdot x = 63.546 - 64.9278\] \[1.9982 \cdot x = 1.3818\]
\[x = \] $\dfrac{{1.38181}}{{0.9982}}$
\[x = {\text{ }}0.69152\]
Step 4: Find percent abundance
This implies that the percent abundances of the two isotopes will be
\[69.152\% \]---->\[^{63}Cu\]
\[30.848\% \]------.\[^{65}Cu\]
Note:
If a mass spectrum of the component was given, the relative rate isotope abundances are generally introduced as a vertical bar graph. The all-out may look as though it exceeds \[100\% ,\] however, that is because the mass spectrum works with relative rate isotope abundances.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)