Question

A stationary free electron in a gas is struck by an X-ray with an energy of \(175.37 \mathrm{eV}\), which experiences a decrease in energy of \(28.52 \mathrm{meV}\). How fast is the electron moving after the interaction with the X-ray?

Short answer

Answer: The electron's velocity after interacting with the X-ray photon is approximately \(2.8942\times10^6\,\frac{\text{m}}{\text{s}}\).

Step-by-step solution

Convert given energy values to Joules

First, we need to convert the given energy values (175.37 eV and 28.52 meV) to Joules using the conversion factor 1 eV = \(1.6\times10^{-19}\) J. Initial Energy of X-ray photon in Joules: \(E_{initial} = 175.37\,\text{eV} \times 1.6\times10^{-19}\,\frac{\text{J}}{\text{eV}} = 2.80592\times10^{-17}\,\text{J}\) Decrease in Energy of X-ray photon in Joules: \(\Delta E = 28.52\,\text{meV}\times10^{-3}\,\frac{\text{eV}}{\text{meV}}\times1.6\times10^{-19}\,\frac{\text{J}}{\text{eV}} = 4.5632\times10^{-20}\,\text{J}\)

Determine the energy transferred to the electron

Use the principle of conservation of energy to find the energy transferred to the electron. The energy given to the electron (\(E_{electron}\)) is the difference between the initial photon energy and the decrease in energy: \(E_{electron}=E_{initial}-\Delta E = 2.80592\times10^{-17}\,\text{J} - 4.5632\times10^{-20}\,\text{J} = 2.760256\times10^{-17}\,\text{J}.\)

Use the kinetic energy formula to find the velocity of the electron

We can use the kinetic energy formula to find the velocity of the electron. The kinetic energy of an object is given by the formula \(K = \frac{1}{2}mv^2\), where \(K\) is the energy, \(m\) is the mass, and \(v\) is the velocity. First, we need to know the mass of the electron (\(m\)). The mass of an electron is about \(9.109\times10^{-31}\,\text{kg}\). We can rearrange the kinetic energy formula to solve for the velocity: \(v = \sqrt{\frac{2K}{m}}\) Now substitute the values for \(K\) and \(m\): \(v = \sqrt{\frac{2 \times 2.760256\times10^{-17}\,\text{J}}{9.109\times10^{-31}\,\text{kg}}} = \sqrt{\frac{5.520512\times10^{-17}\,\text{J}}{9.109\times10^{-31}\,\text{kg}}} \approx 2.8942\times10^6\,\frac{\text{m}}{\text{s}}\)

Write the final answer

After the interaction with the X-ray photon, the electron moves with a velocity of approximately \(2.8942\times10^6\,\frac{\text{m}}{\text{s}}\).