Animated Solution for Physics - Atoms and Nuclei: The $K_{\alpha }$ X-ray emission line of tungsten occurs at $\lambda =0.021\text{ nm}$. The energy difference between $K$ and $L$ levels in this atom is about

When we bombard a heavy target like tungsten with high-energy electrons, we can knock out an electron from one of its innermost shells, such as the $K$-shell ($n=1$). This creates a vacancy, making the atom highly unstable. To restore stability, an electron from a higher energy level, like the $L$-shell ($n=2$), quickly drops down to fill the hole.

As this electron falls into the deeper potential well, it must shed its excess energy. It does this by emitting a single, highly energetic photon. Because this transition is specifically from the $L$-shell to the $K$-shell, the emitted photon forms what we call the $K_{\alpha }$ characteristic X-ray line.

The energy of this emitted photon is exactly equal to the energy difference between the two shells involved in the transition. We can express this mathematically using the Planck-Einstein relation:

Here, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the emitted X-ray.

While we could plug in the standard SI values for $h$ and $c$, there is a much faster shortcut widely used in modern physics. If we express the wavelength $\lambda$ in Angstroms ($\text{A˚}$), the energy $\Delta E$ in electron-volts (eV) is given by:

Our given wavelength is $\lambda =0.021\text{ nm}$. Since $1\text{ nm}=10\text{ A˚}$, we can easily convert this:

Now, we simply substitute our wavelength into the shortcut formula:

Looking at our options, they are given in MeV and keV. Let's convert our answer to kilo electron-volts (keV) by dividing by $1000$:

Rounding to the nearest integer, we get $59\text{ keV}$, which perfectly matches option (c).