Chapter 6: Problem 5
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?
Short Answer
Expert verified
Answer: Yes, X-rays with a wavelength of 85 nm have sufficient energy to ionize rubidium atoms.
Step by step solution
01
Calculate the energy of a single X-ray photon
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
02
Convert the ionization energy to energy per atom
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
03
Compare the energy of an X-ray photon to the ionization energy of rubidium
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
X-ray photon energy
X-ray photon energy refers to the energy carried by an X-ray photon, which can be used to understand how X-rays interact with atoms. In this exercise, we determine the energy of an X-ray photon given its wavelength. X-rays are a form of electromagnetic radiation, just like visible light, but they have higher energy and shorter wavelengths.
- The energy of these photons is crucial when considering their ability to ionize atoms.
- Ionization involves the removal of an electron, which requires overcoming the atom’s ionization energy.
Planck's equation
Planck's equation is a fundamental relation used to connect the energy of a photon to its frequency. The equation is represented as \( E = h \cdot f \), where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant, approximately \(6.626 \times 10^{-34} \) J/Hz, and
- \( f \) is the frequency of the photon.
Ionization of rubidium
Ionization of rubidium involves removing an electron from a neutral rubidium atom, which requires a specific amount of energy known as the ionization energy. For rubidium, this energy is approximately \( 403 \mathrm{~kJ/mol} \). This quantity needs to be converted into energy per atom for individual assessment. To convert the molar ionization energy to a per-atom basis:
- We first convert \( 403 \mathrm{~kJ/mol} \) to joules: \( 403 \mathrm{~kJ/mol} = 403 \times 1000 \mathrm{~J/mol} \).
- Then divide by Avogadro’s number, \( 6.022 \times 10^{23} \mathrm{atoms/mol} \).
Energy conversion
Energy conversion in this context refers to converting given energetic values into suitable units for comparisons. Here, we need to convert the energies involved:
- The ionization energy of rubidium is given in \( \mathrm{~kJ/mol} \), which we convert to joules to match the units of photonic energy.
- The X-ray photon energy, calculated using Planck’s equation, is already in joules, making direct comparisons straightforward.