Animated Solution for Physics - Atoms and Nuclei: X-rays are incident on a target metal atom having 30 neutrons. The ratio of atomic radius of the target atom and ${}_2^4\text{He}$ is $(14{)}^{1/3}$. (a) Find the mass number of target atom. (b) Find the frequency of $K_{\alpha }$ line emitted by this metal. ($R=1.1\times 10^7{\text{ m}}^{-1},c=3\times 10^8\text{ m/s}$)

The journey through modern physics often requires us to connect seemingly disparate concepts to reveal the hidden properties of matter. In this fascinating problem, we are tasked with identifying an unknown target metal atom and then determining the frequency of the X-rays it emits. This requires a beautiful synthesis of nuclear physics and atomic spectra. Let's dive in!

The problem begins by giving us a clue about the size of the unknown target atom's nucleus. It states that the ratio of the "atomic radius" of the target atom to that of a Helium-4 nucleus is $(14{)}^{1/3}$.

A quick note on terminology: The problem uses the phrase "atomic radius," but the mathematical dependence provided—the cube root—is the hallmark of the nuclear radius. The radius $r$ of a nucleus is empirically found to be directly proportional to the cube root of its mass number $A$:

where $r_0$ is a constant (approximately $1.2\times 10^{-15}$ m).

By setting up a ratio for the target atom (let's call it atom 2) and the Helium nucleus (atom 1), we can write:

We know that the mass number of the Helium nucleus (${}_2^4\text{He}$) is $A_1=4$. Substituting the given ratio into our equation, we get:

Cubing both sides to eliminate the fractional exponent, we find:

We have successfully determined the mass number of the target atom! But to understand its chemical identity and its X-ray emission properties, we need its atomic number, $Z$. The problem tells us that the target atom has $30$ neutrons. Since the mass number is the sum of protons and neutrons ($A=Z+n$), we can easily find $Z$:

An atomic number of $26$ corresponds to Iron (Fe). Our unknown target is Iron!

Now that we know the target is Iron ($Z=26$), we need to find the frequency of the $K_{\alpha }$ line emitted by this metal when bombarded with X-rays.

The $K_{\alpha }$ line is produced when an electron from the L-shell ($n=2$) transitions down to fill a vacancy in the K-shell ($n=1$). The frequency of this emitted X-ray photon is governed by Moseley's Law, which is a modification of the Rydberg formula that accounts for the shielding effect of the remaining inner-shell electrons.

The formula for the frequency $f$ is:

For the $K_{\alpha }$ line, the electron transitioning from the L-shell "sees" a nuclear charge that is shielded by the one remaining electron in the K-shell. Therefore, the screening constant $b$ is approximately $1$.

Substituting $b=1$, $n_1=1$, and $n_2=2$, the formula simplifies to:

We are given the Rydberg constant $R=1.1\times 10^7{\text{ m}}^{-1}$ and the speed of light $c=3\times 10^8\text{ m/s}$. Let's plug in all our values, remembering that $Z=26$:

To express this in standard scientific notation, we shift the decimal point:

And there we have it! By seamlessly connecting the empirical formula for nuclear size with Moseley's law for X-ray spectra, we've completely unraveled the mysteries of the target atom.