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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 (位): f = c / 位 where c is the speed of light (approximately \(3 * 10^8 \mathrm{~m/s}\)) and 位 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 鈮� \(3.529 * 10^{16} \mathrm{~Hz}\) 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 鈮� 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 鈮� 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 鈮� 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鈥檚 ionization energy.
Understanding X-ray photon energy helps assess whether the X-ray radiation can cause ionization in different substances, like rubidium in this instance.
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.
To find the frequency, we use the relation between frequency \( f \) and wavelength \( \lambda \): \( f = \frac{c}{\lambda}\), where \( c \) is the speed of light, approximately \(3 \times 10^8\) m/s.Planck鈥檚 equation plays a vital role in determining whether a given X-ray has enough energy to ionize atoms by giving insight into its energetic capacity.
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鈥檚 number, \( 6.022 \times 10^{23} \mathrm{atoms/mol} \).
This conversion allows us to understand and compare how individual photons interact with individual atoms, which is essential to determining ionization feasibility.
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鈥檚 equation, is already in joules, making direct comparisons straightforward.
Understanding energy conversion is critical, as it allows us to derive meaningful relationships between different forms of energy and make accurate scientific judgments, such as determining if an X-ray photon can ionize a rubidium atom.

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