Animated Solution for Physics - Atoms and Nuclei: If ${\lambda }_{\text{Cu}}$ is the wavelength of $K_{\alpha }$, X-ray line of copper (atomic number 29) and ${\lambda }_{\text{MO}}$ is the wavelength of the $K_{\alpha }$, X-ray line of molybdenum (atomic number 42), then the ratio ${\lambda }_{\text{Cu}}/{\lambda }_{\text{Mo}}$ is close to

The problem asks us to find the ratio of the wavelengths of the $K_{\alpha }$ X-ray lines for Copper (Cu) and Molybdenum (Mo). To understand this, we first need to visualize what a $K_{\alpha }$ transition is.

Imagine an atom where an inner-shell electron has been knocked out, leaving a vacancy in the K-shell ($n=1$). An electron from the L-shell ($n=2$) jumps down to fill this vacancy. The energy difference between these two shells is released as an X-ray photon. This specific emission is called the $K_{\alpha }$ line.

To find the wavelength of this emitted X-ray, we use Moseley's Law, which modifies the Rydberg formula to account for the shielding effect of the remaining inner electrons. The formula is given by:

Here, $R$ is the Rydberg constant, $Z$ is the atomic number, and $b$ is the screening constant. For a $K_{\alpha }$ transition, the jumping electron is shielded by the single remaining electron in the K-shell, so $b\approx 1$. The transition is from $n_2=2$ to $n_1=1$.

Since $R$, $n_1$, and $n_2$ are the same for the $K_{\alpha }$ transition in any element, all these terms become a single constant. This leaves us with a beautiful, simple proportionality:

This means the wavelength $\lambda$ is inversely proportional to $(Z-1{)}^2$.

We need the ratio of the wavelength of Copper to Molybdenum, $\frac{{\lambda }_{\text{Cu}}}{{\lambda }_{\text{Mo}}}$. Because of the inverse relationship, the atomic number of Molybdenum will appear in the numerator:

Now, we simply substitute the given atomic numbers. For Copper, $Z_{\text{Cu}}=29$, and for Molybdenum, $Z_{\text{Mo}}=42$.

Calculating the squares, we get $41^2=1681$ and $28^2=784$.