Animated Solution for Physics - Atoms and Nuclei: The wavelength of the characteristic X-ray $K_{\alpha }$ line emitted by a hydrogen like element is $0.32\text{A˚}$. The wavelength of the $K_{\beta }$ line emitted by the same element will be ........ .

Imagine you are looking at the energy levels of a hydrogen-like atom. When an electron jumps from a higher energy level to a lower one, it emits a photon. In the context of characteristic X-rays, these transitions have specific names. The $K_{\alpha }$ line is emitted when an electron jumps from the L-shell ($n=2$) to the K-shell ($n=1$). Similarly, the $K_{\beta }$ line is emitted when an electron jumps from the M-shell ($n=3$) to the K-shell ($n=1$).

Moseley's Law

To find the wavelength of these emitted X-rays, we rely on a combination of Moseley's Law and the Rydberg formula. The formula is given by:

Here, $R$ is the Rydberg constant, $Z$ is the atomic number, and $b$ is the screening constant. For the K-series transitions, the screening constant $b$ is approximately $1$. Since we are dealing with the same element for both transitions, the term $R(Z-1{)}^2$ remains constant. This means the wavelength $\lambda$ is inversely proportional to the difference of the inverse squares of the principal quantum numbers:

By taking the ratio of the wavelengths of the $K_{\beta }$ and $K_{\alpha }$ lines, we can eliminate the constant terms. The ratio is purely dependent on the quantum numbers:

Let's calculate the values inside the brackets. For the $K_{\alpha }$ transition, we have $1-\frac{1}{4}=\frac{3}{4}$. For the $K_{\beta }$ transition, we have $1-\frac{1}{9}=\frac{8}{9}$. Substituting these back into our ratio gives:

Simplifying the complex fraction, we get $\frac{27}{32}$. Now, we can easily find the wavelength of the $K_{\beta }$ line by multiplying this fraction by the given wavelength of the $K_{\alpha }$ line ($0.32\text{A˚}$):

Notice that the $K_{\beta }$ transition involves a larger energy difference than the $K_{\alpha }$ transition. Because energy is inversely proportional to wavelength, it makes perfect sense that the $K_{\beta }$ wavelength is shorter than the $K_{\alpha }$ wavelength. Always keep this physical intuition in mind to cross-check your answers!