Problem 38 - University Physics with Modern Physics

Question

The energies for an electron in the \(K, L,\) and \(M\) shells of the tungsten atom are \(-69,500 \mathrm{eV},-12,000 \mathrm{eV},\) and \(-2200 \mathrm{eV}\) respectively. Calculate the wavelengths of the \(K_{\alpha}\) and \(K_{\beta} \mathrm{x}\) rays of tungsten.

Step-by-Step Solution

Answer

\(\lambda_{K_{\alpha}} = 2.164 \times 10^{-11} \text{ m}\), \(\lambda_{K_{\beta}} = 1.841 \times 10^{-11} \text{ m}\).

Step 1: Understanding the Problem

We need to calculate the wavelengths of two X-ray transitions, \(K_{\alpha}\) and \(K_{\beta}\), for tungsten. The \(K_{\alpha}\) transition involves an electron dropping from the \(L\) to the \(K\) shell, and the \(K_{\beta}\) transition involves an electron dropping from the \(M\) to the \(K\) shell.

Step 2: Energy Difference for \(K_{\alpha}\) Transition

The energy difference for \(K_{\alpha}\) is the energy difference between the \(L\) and \(K\) shells. Calculate: \( \Delta E = E_L - E_K = (-12,000 \text{ eV}) - (-69,500 \text{ eV}) = 57,500 \text{ eV} \).

Step 3: Energy Difference for \(K_{\beta}\) Transition

The energy difference for \(K_{\beta}\) is the energy difference between the \(M\) and \(K\) shells. Calculate: \( \Delta E = E_M - E_K = (-2200 \text{ eV}) - (-69,500 \text{ eV}) = 67,300 \text{ eV} \).

Step 4: Using Energy-Wavelength Relationship for \(K_{\alpha}\)

Use the energy-wavelength relationship \(E = \frac{hc}{\lambda}\) to find \(\lambda\). Substitute \(\Delta E = 57,500 \text{ eV}\), \(h = 4.135667696 \times 10^{-15} \text{ eV} \cdot \text{s} \), and \(c = 2.998 \times 10^8 \text{ m/s}\). Rearrange to find \(\lambda = \frac{hc}{\Delta E}\).

Step 5: Calculating \(\lambda_{K_{\alpha}}\)

Calculate \(\lambda_{K_{\alpha}}\) with \(\lambda = \frac{4.135667696 \times 10^{-15} \times 2.998 \times 10^8}{57,500} = 2.164 \times 10^{-11} \text{ m}\).

Step 6: Using Energy-Wavelength Relationship for \(K_{\beta}\)

Similarly, use the same formula for the \(K_{\beta}\) transition: \(\lambda = \frac{hc}{\Delta E}\), where \(\Delta E = 67,300 \text{ eV}\).

Step 7: Calculating \(\lambda_{K_{\beta}}\)

Calculate \(\lambda_{K_{\beta}}\) with \(\lambda = \frac{4.135667696 \times 10^{-15} \times 2.998 \times 10^8}{67,300} = 1.841 \times 10^{-11} \text{ m}\).

Step 8: Final Result Comparison

Now we have the wavelength for each transition: \(\lambda_{K_{\alpha}} = 2.164 \times 10^{-11} \text{ m}\) and \(\lambda_{K_{\beta}} = 1.841 \times 10^{-11} \text{ m}\). These wavelengths correspond to the respective transition energies.

Key Concepts

Energy Levels

In atoms, electrons are arranged in energy levels or shells, named as K, L, M, and so on. These levels represent the electron's energy state, with the K-shell being the closest to the nucleus and having the lowest energy level among them. In our tungsten example:

• K-shell energy is -69,500 eV
• L-shell energy is -12,000 eV
• M-shell energy is -2,200 eV

The negative sign indicates that these energy levels are bound states, meaning the electrons are confined within the atom's electric potential well. Electrons in outer shells have higher energy (less negative) because they are further from the nucleus. When electrons transition between these levels, they either absorb or release energy in the form of photons. This process is fundamental in understanding X-ray transitions, like the \( K_{\alpha} \) and \( K_{\beta} \).

Wavelength Calculation

To calculate the wavelength of X-rays resulting from electron transitions, we first need the energy difference between the initial and final states. For instance, the \( K_{\alpha} \) transition in tungsten involves an electron moving from the L-shell to the K-shell, producing a photon with energy equal to this difference.When we calculate the difference for \( K_{\alpha} \):\[ \Delta E = E_L - E_K = (-12,000 \text{ eV}) - (-69,500 \text{ eV}) = 57,500 \text{ eV} \] For the \( K_{\beta} \) transition:\[ \Delta E = E_M - E_K = (-2,200 \text{ eV}) - (-69,500 \text{ eV}) = 67,300 \text{ eV} \] These energy differences allow us to use the energy-wavelength formula \( E = \frac{hc}{\lambda} \) to find the corresponding wavelengths, a crucial step in understanding how X-rays work and are measured. This, in turn, helps in applications like medical imaging and material analysis.

Energy-Wavelength Relationship

The relationship between energy and wavelength is one of the foundations of understanding X-rays and many other spectroscopic techniques. This relationship is mathematically defined by the formula:\[ E = \frac{hc}{\lambda} \]Where:

• \( E \) is the energy of the photon (in \( ext{eV} \))
• \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \text{ eV}\cdot\text{s} \))
• \( c \) is the speed of light (\( 2.998 \times 10^8 \text{ m/s} \))

To find the wavelength \( \lambda \), you rearrange the formula to \( \lambda = \frac{hc}{\Delta E} \). Applying this to our example, for \( K_{\alpha} \) transition: \[ \lambda_{K_{\alpha}} = \frac{4.135667696 \times 10^{-15} \times 2.998 \times 10^8}{57,500} = 2.164 \times 10^{-11} \text{ m} \] And for \( K_{\beta} \) transition: \[ \lambda_{K_{\beta}} = \frac{4.135667696 \times 10^{-15} \times 2.998 \times 10^8}{67,300} = 1.841 \times 10^{-11} \text{ m} \] This energy-wavelength relationship allows us to convert measured energies into wavelengths directly. Both \( K_{\alpha} \) and \( K_{\beta} \) lines are critical for applications like X-ray fluorescence analysis.