3) give the relative abundance of the two isotopes of nitrogen when n-14 hos o 14.0031 amu and n-15 has a mass of 15.0001 amu

The relative abundance of N-14 = 99.64 %.

The relative abundance of N-15 = 0.361 %.

Explanation:

The average atomic mass of an element can be calculated as the summation of the atomic mass units of each isotope multiplied by its (abundance % / 100).

∴ The average atomic mass of nitrogen (N) = [(atomic mass of N-14)(Abundance / 100) + (atomic mass of N-15)(Abundance / 100).

The average atomic mass of nitrogen (N) = [(atomic mass of N-14)(x)] + (atomic mass of N-15)(1-x)].

The average atomic mass of nitrogen (N) = 14.0067 amu.

Atomic mass of N-14 = 14.0031 amu.

x is (the abundance of N-14 / 100).

Atomic mass of N-15 = 15.0001 amu.

(1-x) is (the abundance of N-15 / 100).

The average atomic mass of nitrogen (N) = [(atomic mass of N-14)(x)] + (atomic mass of N-15)(1-x)].

14.0067 amu = [(14.0031 amu)(x)] + [(15.0001 amu)(1-x)]

14.0067 = 14.0031 x + 15.0001 - 15.0001 x.

(14.0067) - (15.0001) = (14.0031 x) - (15.0001 x).

- 0.9934 = - 0.997 x.

x = (- 0.9934) / (- 0.997) = 0.9964.

x is the abundance of N-14 / 100 = 0.9964.The relative abundance of N-14 = 99.64 %.The abundance of N-15 / 100 = (1-x) = 1 - 0.9963 = 3.61 x 10⁻³.The relative abundance of N-15 = 0.361 %.

Determine the  wavelength of light emitted when an electron  in an li2+ ion  makes a transition from an orbital with n  =  6 to an orbital with n  = 5.

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