Animated Solution for Physics - Atoms and Nuclei: Characteristic X-rays of frequency $4.2\times 10^{18}$ Hz are produced when transitions from L-shell to K-shell take place in a certain target material. Use Mosley's law to determine the atomic number of the target material. Given Rydberg's constant $R=1.1\times 10^7{\text{ m}}^{-1}$.

An epic journey into the heart of the atom! Imagine you are a quantum detective, and the only clue you have is a flash of invisible light—an X-ray photon. By simply measuring the frequency of this light, you can deduce exactly how many protons are hiding in the nucleus of the atom that emitted it. This is the magic of Moseley's Law. Let's break down this beautiful problem step by step.

The problem states that an electron transitions from the L-shell to the K-shell. In the language of quantum mechanics, the shells are numbered starting from the nucleus outwards. The K-shell is the first orbit ($n=1$), and the L-shell is the second orbit ($n=2$).

When an electron falls from a higher energy state ($n=2$) to a lower energy state ($n=1$), it must shed its excess energy. It does this by emitting a photon. Because this energy gap is massive in heavy elements, the emitted photon is a high-energy X-ray, specifically known as the $K_{\alpha }$ line.

To connect the frequency of this X-ray to the atomic number $Z$, we use Moseley's Law. This law is a brilliant adaptation of the Bohr model for multi-electron atoms. The formula is:

Here, $R$ is the Rydberg constant, $c$ is the speed of light, and $b$ is the screening constant. For a K-series transition, the electron falling into the K-shell "sees" the nucleus, but its view is slightly blocked by the one electron that is already sitting in the K-shell. This single electron shields exactly one unit of positive charge, making the effective nuclear charge $(Z-1)e$. Thus, for $K_{\alpha }$ transitions, $b=1$.

Let's substitute the values given in the problem into our master equation:

First, let's simplify the constants on the right side. The product of the Rydberg constant and the speed of light is:

Next, we evaluate the orbital fraction:

Now, our equation looks much less intimidating:

Multiplying the constants gives:

To find the atomic number, we need to isolate $(Z-1{)}^2$. We do this by dividing the left side by our combined constant:

To make the division easier, let's borrow $10^3$ from the numerator's power of ten:

Calculating this division gives:

We are at the finish line! Taking the square root of both sides:

Since an atom cannot have a fraction of a proton, the atomic number must be an integer. Rounding to the nearest whole number, we get $Z=42$.

The target material is Molybdenum! It is incredibly profound that a simple algebraic calculation allows us to peer into the quantum structure of matter.