X-rays are produced in an x-ray tube by electrons accelerated through an electric potential difference of 50 kV. Let${\mathbf{K}}_{\mathbf{0}}$ be the kinetic energy of an electron at the end of the acceleration. The electron collides with a target nucleus (assume the nucleus remains stationary) and then has kinetic energy ${\mathbf{K}}_1\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}{\mathbf{K}}_{\mathbf{0}}$. (a) What wavelength is associated with the photon that is emitted? The electron collides with another target nucleus (assume it, too, remains stationary) and then has kinetic energy ${\mathbf{K}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{500}{\mathbf{K}}_{\mathbf{1}}$. (b) What wavelength is associated with the photon that is emitted?

Accelerating potential difference of the electrons,${\mathrm{K}}_0=50\mathrm{kV}$

The kinetic energy of the first target, ${\mathrm{K}}_1=0.5{\mathrm{K}}_0$

The kinetic energy of the second target,${\mathrm{K}}_2=0.5{\mathrm{K}}_1$

Consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{15}\times 6.63\times {10}^{-34}\right)\mathrm{keV}.\mathrm{s} \\ =41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^{12}\right)\mathrm{pm}/\mathrm{s} \\ =3\times {10}^{20}\mathrm{pm}/\mathrm{s} \end{gathered}$$

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Use the concept of Planck's relation to finding the wavelength for each condition of the collision by substituting the energy values of the kinetic energies of the target nucleus that affects the initial energy of the electron.

Formula:

The energy of the photon due to Planck’s relation,

$∆E=\frac{hc}{\lambda }$ ….. (1)

Here,$∆E$ is the energy of the photon,h is the Plank’s constant, and c is the speed of light.

Using the given condition and initial energy, the kinetic energy of the target nucleus can be given as:

$$\begin{gathered} {\mathrm{K}}_1=0.5\left(50\mathrm{kV}\right) \\ =25\mathrm{kV} \end{gathered}$$

Now, the energy of the emitted photon can be given as follows:

$$\begin{gathered} \mathrm{E}=\left(50-25\right)\mathrm{kV} \\ =25\mathrm{kV} \end{gathered}$$

Thus, the wavelength associated with the emitted photon due to change in energy after collision with the first target can be given using the above value in equation (1) as follows:

$$\begin{gathered} \mathrm{\lambda }=\frac{\left(41.384\times {10}^{-19}\mathrm{keV}.\mathrm{s}\right)\left(3\times {10}^{20}\mathrm{pm}/\mathrm{s}\right)}{25\times {10}^3\mathrm{eV}} \\ =\frac{1240\mathrm{keV}.\mathrm{nm}}{25\times {10}^3\mathrm{eV}} \\ =4.96\times {10}^{-2}\mathrm{nm} \\ =49.6\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is 49.6 pm.

Using the given condition and initial energy, the kinetic energy of the target nucleus can be given as:

$$\begin{gathered} {\mathrm{K}}_2=0.5\left(25\mathrm{kV}\right) \\ =12.5\mathrm{kV} \end{gathered}$$

Now, the energy of the emitted photon produced in the second and third collision is same that is:

E=12.5 kV

Thus, the wavelength associated with the emitted photon due to change in energy after collision with the first target can be given using the above value in equation (1) as follows:

$$\begin{gathered} \mathrm{\lambda }=\frac{1240\mathrm{keV}.\mathrm{nm}}{12.5\times {10}^3\mathrm{eV}} \\ =9.92\times {10}^{-2}\mathrm{nm} \\ =99.2\mathrm{pm} \end{gathered}$$

Hence, the value of the wavelength is 99.2 pm.