Question

the first ionization energy, e, of a boron atom is 8.01 x 10² kj/mol. what is the wavelength, in nm, that is just sufficient to ionize a boron atom? report your answer using 3 significant figures. short answer type answer here

Explanation:

Step1: Calculate energy per atom

Given ionization energy \(E = 8.01\times10^{2}\text{ kJ/mol}\).
Convert to energy per atom:
\(E_{\text{atom}}=\frac{8.01\times 10^{2}\times10^{3}\text{ J/mol}}{6.022\times 10^{23}\text{ atoms/mol}}\)
\(E_{\text{atom}}=\frac{8.01\times 10^{5}}{6.022\times 10^{23}}\text{ J/atom}\approx1.33\times 10^{-18}\text{ J/atom}\)

Step2: Use the formula \(E = h

u=\frac{hc}{\lambda}\)
Rearrange for \(\lambda\): \(\lambda=\frac{hc}{E}\)
where \(h = 6.626\times 10^{-34}\text{ J s}\), \(c = 3\times 10^{8}\text{ m/s}\), and \(E = 1.33\times 10^{-18}\text{ J}\)
\(\lambda=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{1.33\times 10^{-18}}\text{ m}\)
\(\lambda=\frac{19.878\times 10^{-26}}{1.33\times 10^{-18}}\text{ m}\approx1.49\times 10^{-7}\text{ m}\)

Step3: Convert meters to nanometers

Since \(1\text{ m}=10^{9}\text{ nm}\)
\(\lambda = 1.49\times 10^{-7}\times10^{9}\text{ nm}=149\text{ nm}\)

Answer:

\(149\text{ nm}\)