The ionization energy of rubidium is \(403 \mathrm{~kJ} / \mathrm{mol}\). Do X-rays with a wavelength of \(85 \mathrm{~nm}\) have sufficient energy to ionize rubidium?

To calculate the energy of a single X-ray photon, we will use the Planck's equation: E = h * f where E is the energy of the photon, h is the Planck's constant, and f is the frequency of the X-ray. First, we need to find the frequency (f) from the given wavelength (λ): f = c / λ where c is the speed of light (approximately \(3 * 10^8 \mathrm{~m/s}\)) and λ is the wavelength in meters (given as 85 nm, or \(85 * 10^{-9} \mathrm{~m}\) in SI units). Now we can calculate the frequency f: f = \(3 * 10^8 \mathrm{~m/s}\) / (\(85 * 10^{-9} \mathrm{~m}\)) f ≈ \(3.529 * 10^{16} \mathrm{~Hz}\) Next, we can find the energy of a single photon using the Planck's equation and the Planck's constant (approximately \(6.626 * 10^{-34} \mathrm{~J/Hz}\)): E = h * f E = (\(6.626 * 10^{-34} \mathrm{~J/Hz}\)) * (\(3.529 * 10^{16} \mathrm{~Hz}\)) E ≈ 2.336 * 10^-17 J

We are given the ionization energy of rubidium in kJ/mol. To compare this to the energy of a single X-ray photon, we need to convert it to energy per atom. We do this by dividing the ionization energy by Avogadro's number (approximately \(6.022 * 10^{23} \mathrm{atoms/mol}\)): Ionization energy per atom = (Ionization energy in kJ/mol) / (Avogadro's number) (1 kJ = 1000 J) Ionization energy per atom = (403 kJ/mol * 1000 J/1 kJ) / (\(6.022 * 10^{23} \mathrm{atoms/mol}\)) Ionization energy per atom ≈ 6.694 * 10^-19 J

Now we can compare the energy of a single X-ray photon (E ≈ 2.336 * 10^-17 J) to the ionization energy of rubidium per atom (6.694 * 10^-19 J). Since the energy of an X-ray photon is greater than the ionization energy of rubidium per atom: 2.336 * 10^-17 J > 6.694 * 10^-19 J We can conclude that X-rays with a wavelength of 85 nm have sufficient energy to ionize rubidium atoms.