To find the abundance and mass of the other isotope of vanadium, we can use the concept of weighted averages based on the relative abundance of the isotopes.
A. Finding the abundance of the other isotope
Let's denote:
- The mass of the first isotope \( m_1 = 50.9440 , \text{amu} \)
- The abundance of the first isotope \( a_1 = 99.75% = 0.9975 \)
- The abundance of the second isotope \( a_2 \)
- We know that the total abundance must equal 100%, so:
\[ a_1 + a_2 = 1 \]
This means:
\[ a_2 = 1 - a_1 = 1 - 0.9975 = 0.0025 , \text{or} , 0.25% \]
So, the abundance of the other isotope is 0.25%.
B. Finding the mass of the other isotope
We can set up the equation for the average atomic mass based on the isotopes and their abundances:
\[ \text{Average atomic mass} = (m_1 \cdot a_1) + (m_2 \cdot a_2) \]
Where:
- The average atomic mass of vanadium is \( 50.9415 , \text{amu} \)
- Rearranging for \( m_2 \) (mass of the second isotope) gives us:
\[ 50.9415 = (50.9440 \cdot 0.9975) + (m_2 \cdot 0.0025) \]
Calculating \( (50.9440 \cdot 0.9975) \):
\[ 50.9440 \cdot 0.9975 \approx 50.8408 \]
Now we plug this into our equation:
\[ 50.9415 = 50.8408 + (m_2 \cdot 0.0025) \]
Subtract \( 50.8408 \) from both sides:
\[ 50.9415 - 50.8408 = m_2 \cdot 0.0025 \]
\[ 0.1007 = m_2 \cdot 0.0025 \]
Now, solve for \( m_2 \):
\[ m_2 = \frac{0.1007}{0.0025} \approx 40.28 , \text{amu} \]
Thus, the mass of the other isotope is approximately 40.28 amu.
Summary:
A. The abundance of the other isotope is 0.25%.
B. The mass of the other isotope is approximately 40.28 amu.