Answer:
2.40 × 10² nm.
Explanation:
We are given that the the first ionization of a sodium atom is 5.00 × 10² kJ/mol and we want to determine the wavelength of light (in nm) that is sufficient to ionize a sodium atom.
We will first find the ionization energy required for one sodium atom using molar conversions:
[tex]\displaystyle 1 \text{ atom Na} \cdot \frac{5.00 \times 10^2\text{ kJ}}{\text{mol Na}} \cdot \frac{1\text{ mol Na}}{6.02\times 10^{23} \text{ atom Na}} = 8.31 \times 10^{-22} \text{ kJ}[/tex]
From the formula:
[tex]\displaystyle E = hv[/tex]
Where h is Planck's constant, substitute and solve for frequency v:
[tex]\displaystyle \begin{aligned} \left(8.31 \times 10^{-22} \text{ kJ}\right) &= (6.626\times 10^{-34} \text{J$\cdot$s})\left(\frac{1 \text{ kJ}}{1000 \text{ J}}\right) v \\ \\ v & = 1.25\times 10^{15} \text{ s}^{-1}\end{aligned}[/tex]
From the relationship between frequency and wavelength, solve for wavelength:
[tex]\displaystyle \begin{aligned} c & = \lambda\nu \\ \\ \lambda & = \frac{c}{v} \\ \\ & = \frac{(3.00 \times 10^8 \text{ ms}^{-1})}{(1.25 \times 10^{15}\text{ s}^{-1})}\cdot \left( \frac{1\times 10^9\text{ nm}}{1 \text{ m}}\right)\\ \\ &=2.40\times 10^2 \text{ nm} \end{aligned}[/tex]
In conclusion, the wavelength of light necessary to ionize a sodium ion is about 2.40 × 10² nm.
