Animated Solution for Physics - Atoms and Nuclei: The wavelength of $K_{\alpha }$, X-rays produced by an X-ray tube is $0.76\text{ A˚}$. The atomic number of the anode material of the tube is ......... .

Have you ever wondered how scientists can identify the exact elemental composition of a mysterious metal just by shining high-energy electrons at it? The secret lies in the beautiful, invisible light it emits: X-rays.

In this thrilling problem, we are going to play the role of a quantum detective. We are given the wavelength of a specific X-ray—the $K_{\alpha }$ line—and our mission is to uncover the identity of the atom that emitted it.

Imagine an atom as a bustling city with different orbital highways. The innermost highway is the K-shell ($n=1$), and the next one out is the L-shell ($n=2$).

When a high-speed electron from an X-ray tube crashes into the anode target, it can violently knock out an electron from the K-shell, leaving behind a gaping "hole" or vacancy. The atom hates this instability! To fix it, an electron from the higher L-shell immediately jumps down to fill the void.

Because the L-shell has higher energy than the K-shell, this jumping electron must shed its excess energy. It does so by firing off a high-energy photon. This specific photon is what we call a $K_{\alpha }$ X-ray.

To connect the wavelength of this X-ray to the identity of the atom, we use a brilliant piece of physics called Moseley's Law. It is essentially the Rydberg formula, but upgraded for multi-electron atoms.

Here is the genius part: the screening constant $b$. When the L-shell electron looks down at the nucleus, it doesn't feel the full pull of the atomic number $Z$. Why? Because there is still one electron left in the K-shell (remember, only one was knocked out!). This lone electron acts like a shield, blocking exactly one unit of positive charge. Therefore, for $K_{\alpha }$ transitions, the effective nuclear charge is $(Z-1)$, meaning $b=1$.

Let's plug in the quantum numbers for our $K_{\alpha }$ jump. The electron falls from $n_2=2$ to $n_1=1$.

Simplifying the fractions inside the bracket:

This is our master equation. It beautifully links the macroscopic wavelength $\lambda$ to the microscopic atomic number $Z$.

Our goal is to isolate $Z$. Let's rearrange the equation to solve for $(Z-1{)}^2$:

Taking the square root of both sides gives us:

Now, we bring in the heavy artillery—the numerical values. We are given the Rydberg constant $R=1.097\times 10^7{\text{ m}}^{-1}$ and the wavelength $\lambda =0.76\text{ A˚}$, which is $0.76\times 10^{-10}\text{ m}$.

Let's carefully evaluate the denominator:

Now, divide $4$ by this value:

This number is screaming at us! In the world of physics problems, numbers like $1599.25$ are not random; they are designed to be perfect squares. We can safely approximate this to $1600$.

We are at the finish line. If $Z-1=40$, we simply add $1$ to both sides:

The atomic number is $41$, which corresponds to the element Niobium (Nb).

A Crucial Warning: If you look at the original answer key for this problem, it states the answer is $39$. How did that happen? The author correctly calculated $Z-1=40$, but in a moment of haste, they subtracted $1$ instead of adding it ($40-1=39$).

This is a powerful lesson: never let your guard down on the final algebraic step! You can do all the brilliant quantum mechanics perfectly, but a simple sign error can cost you the points. Always stay sharp until the very end.