Question

the first ionization energy, e, of a lithium atom is 0.860 aj. what is the wavelength of light, in nanometers, that is just sufficient to ionize a lithium atom? values for constants can be found in the chempendix.
λ = nm

Explanation:

Step1: Convert energy unit

Given \(E = 0.860\ \text{aJ}\). Since \(1\ \text{aJ}=10^{-18}\ \text{J}\), then \(E=0.860\times 10^{-18}\ \text{J}\).

Step2: Use the formula \(E = h

u=\frac{hc}{\lambda}\)
where \(h = 6.626\times 10^{-34}\ \text{J}\cdot\text{s}\) (Planck's constant) and \(c = 3\times 10^{8}\ \text{m/s}\) (speed of light).
Rearrange the formula for \(\lambda\): \(\lambda=\frac{hc}{E}\)
Substitute the values: \(\lambda=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{0.860\times 10^{-18}}\)

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Since \(1\ \text{m} = 10^{9}\ \text{nm}\), then \(\lambda=23.1\times10^{-8}\times10^{9}\ \text{nm}\)

Answer:

\(231\ \text{nm}\)