Animated Solution for Physics - Atoms and Nuclei: $K_{\alpha }$ wavelength emitted by an atom of atomic number $Z=11$ is $\lambda$. Find the atomic number for an atom that emits $K_{\alpha }$ radiation with wavelength $4\lambda$

Imagine you are a detective trying to identify an unknown element. You don't have a chemical lab, but you do have an X-ray spectrometer. When you bombard this mystery atom with high-energy electrons, it emits a characteristic $K_{\alpha }$ X-ray. But how does this X-ray tell you the atomic number? This is exactly the puzzle we are solving today!

In 1913, Henry Moseley discovered a beautiful, systematic relationship between the X-ray frequencies emitted by atoms and their atomic numbers. He found that the square root of the frequency $ u$ of the emitted X-ray is directly proportional to the atomic number $Z$ minus a screening constant $b$.

Mathematically, Moseley's Law is written as:

For a $K_{\alpha }$ transition, an electron falls from the $L$ shell ($n=2$) to the $K$ shell ($n=1$). Because there is already one electron sitting in the $K$ shell (since one was knocked out to create the vacancy), it "screens" the nucleus. This makes the effective nuclear charge felt by the falling electron equal to $Z-1$. Thus, for $K_{\alpha }$ lines, the screening constant $b$ is exactly $1$.

We know that frequency $ u$ and wavelength $\lambda$ are inversely related by the speed of light $c$:

Substituting this into Moseley's Law and squaring both sides, we get:

This tells us a profound truth: the reciprocal of the wavelength is directly proportional to the square of $(Z-1)$.

We are given two atoms. The first atom has an atomic number $Z_1=11$ and emits a wavelength ${\lambda }_1=\lambda$. The second atom is our mystery element, $Z_2$, which emits a much longer wavelength, ${\lambda }_2=4\lambda$.

Because the wavelength is longer, the energy of the photon is lower. This immediately tells our intuition that the mystery atom must have a smaller atomic number than $11$. Let's set up a ratio to find the exact value:

Now, we simply plug in the clues we have:

The $\lambda$ terms cancel out perfectly:

Multiplying both sides by $100$:

Taking the square root of both sides (and keeping the positive root since $Z$ must be positive):

And there we have it! The mystery atom has an atomic number of $6$, which means it is Carbon. By simply looking at the wavelength of the light it emitted, we were able to peer into its nucleus and count its protons. That is the true power of physics!