Problem 70 - Fundamentals of Physics

Question

A molybdenum \((Z=42)\) target is bombarded with \(35.0 \mathrm{keV}\) electrons and the \(x\) -ray spectrum of Fig. \(40-13\) results. The \(K_{\beta}\) and \(K_{a}\) wavelengths are \(63.0\) and \(71.0 \mathrm{pm}\), respectively. What photon energy corresponds to the (a) \(K_{\beta}\) and (b) \(K_{\alpha}\) radiation? The two radiations are to be filtered through one of the substances in the following table such that the substance absorbs the \(K_{\beta}\) line more strongly than the \(K_{\alpha}\) line. A substance will absorb radiation \(x_{1}\) more strongly than it absorbs radiation \(x_{2}\) if a photon of \(x_{1}\) has enough energy to eject a \(K\) electron from an atom of the substance but a photon of \(x_{2}\) does not. The table gives the ionization energy of the \(K\) electron in molybdenum and four other substances. Which substance in the table will serve (c) best and (d) second best as the filter? $$ \begin{array}{llllll} \hline & \mathrm{Zr} & \mathrm{Nb} & \mathrm{Mo} & \mathrm{Tc} & \mathrm{Ru} \\\ \hline Z & 40 & 40 & 42 & 43 & 44 \\ E_{K}(\mathrm{keV}) & 18.00 & 18.99 & 20.00 & 21.04 & 22.12 \end{array} $$

Step-by-Step Solution

Answer

(a) \( E_{K_{\beta}} \approx 19.7 \text{ keV} \), (b) \( E_{K_{\alpha}} \approx 17.5 \text{ keV} \). (c) Best filter: Niobium, (d) Second best: Zirconium.

Step 1: Understand the Problem

We need to find the energy of the photons corresponding to the \( K_{\beta} \) and \( K_{\alpha} \) x-ray lines for molybdenum and determine which substance in the table absorbs the \( K_{\beta} \) line more strongly than the \( K_{\alpha} \) line.

Step 2: Convert Wavelengths to Energy

Use the formula \( E = \frac{hc}{\lambda} \) where \( h = 6.63 \times 10^{-34} \text{ J s} \) and \( c = 3 \times 10^8 \text{ m/s} \). Convert pm to meters for the wavelengths.

Step 3: Calculate Energy of K_beta

For \( K_{\beta} \), \( \lambda = 63.0 \text{ pm} = 63.0 \times 10^{-12} \text{ m} \). So, \( E_{K_{\beta}} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{63.0 \times 10^{-12}} \), which gives the energy in joules. Convert this to keV by dividing by \( 1.6 \times 10^{-19} \) and then by 1000.

Step 4: Calculate Energy of K_alpha

For \( K_{\alpha} \), \( \lambda = 71.0 \text{ pm} = 71.0 \times 10^{-12} \text{ m} \). So, \( E_{K_{\alpha}} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{71.0 \times 10^{-12}} \), convert the energy to keV as done for \( K_{\beta} \).

Step 5: Compare Photon Energies with Ionization Energies

Substances will absorb \( K_{\beta} \) more strongly if \( E_{K_{\beta}} > E_K \) and \( E_{K_{\alpha}} < E_K \). Calculate this for each substance in the table.

Step 6: Determine the Best Filter

Check each substance: Determine which substance satisfies \( E_{K_{\alpha}} < E_K < E_{K_{\beta}} \) most strongly.

Key Concepts

Photon Energy

Photon energy is a fundamental concept in X-ray spectroscopy. It describes the energy carried by a photon, which is a packet of electromagnetic radiation. This energy depends on the photon's wavelength or frequency. The formula used to calculate photon energy is

• \[ E = \frac{hc}{\lambda} \]

where:

• \( h \) is Planck's constant \((6.63 \times 10^{-34} \text{ Js})\)
• \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\)
• \( \lambda \) is the wavelength of the photon in meters

Knowing the photon energy is essential for analyzing x-ray spectra in X-ray spectroscopy. This knowledge allows us to understand the interactions between light and matter at a microscopic level. In practical terms, by converting the wavelength given for a certain x-ray line into energy, scientists can identify specific details about the x-ray source, such as its elemental composition.

K-alpha and K-beta Lines

The K-alpha and K-beta lines are specific components of the X-ray emission spectrum of an element. These lines correspond to electronic transitions in atoms that produce x-rays, particularly transitions involving the innermost electron shell (the K-shell).

- **K-alpha Lines**: Occur when an electron transitions from the L-shell (second shell) to the K-shell after being excited or displaced. This transition releases a photon with a certain energy unique to each element. In the context of this exercise, the wavelength for Molybdenum's K-alpha line is given as 71.0 pm.

- **K-beta Lines**: Occur due to electron transitions from higher shells, such as the M-shell to the K-shell. These transitions typically release photons of higher energy than the K-alpha lines. For Molybdenum, the wavelength of the K-beta line is 63.0 pm.

Each element has its unique set of K-alpha and K-beta lines, which help in identifying them and assessing their properties through a process known as X-ray spectroscopy. This makes them a crucial tool for material analysis in scientific research.

Ionization Energy

Ionization energy is the minimum amount of energy required to remove an electron from an atom in its ground state. In the context of x-ray spectroscopy, we often discuss the energy required to eject a K electron, which is an electron in the innermost shell of an atom.

This type of energy is crucial when determining which materials will absorb specific photon energies more effectively. In our exercise, it's important because a material will absorb x-ray radiation more strongly if the photon's energy exceeds the ionization energy of the substance's K electron.

Given the table in the exercise, deciding which substance serves best as a filter involves comparing the calculated energies of the K-beta and K-alpha lines against the ionization energies listed for zirconium (Zr), niobium (Nb), molybdenum (Mo), technetium (Tc), and ruthenium (Ru). The optimal filter will allow the K-alpha line to pass while effectively absorbing the K-beta line, based on this comparison of photon and ionization energies. This understanding highlights the practical applications of x-ray spectroscopy in separating and analyzing different x-ray components.