Question

Determine the rank according to increasing wavelength of$k_{\alpha },k_{\beta }$ and${\text{L}}_{\alpha }.$

Short answer

The ranking of the spectral lines from lowest to highest is $K_{\beta },K_{\alpha }$ and ${\text{L}}_{\alpha }$.

Step-by-step solution

 Step 1: Given data

The given data spectral lines $K_{\beta },K_{\alpha }$ and${\text{L}}_{\alpha }$ .

 Step 2: Concept of Energy

The energy E of the emitted ray is inversely proportional to the wavelength $\lambda$:

$E~\frac{1}{\lambda }$

Determine the electronic configuration

The $K_{\alpha }$ line corresponds to the transition from $n=2$ to $n=1$.

The $K_{\beta }$ line corresponds to the transition from $n=3$ to $n=1$.

The ${\text{L}}_{\alpha }$ line corresponds to the transition from $n=3$ to $n=2$.

The energy E of the emitted ray is inversely proportional to the wavelength$\lambda$ , $E~\frac{1}{\lambda }$.

We know that with increasing n , the energy difference $\Delta E$ becomes smaller.

Thus, the highest energy of the photon corresponds to the $K_{\beta }$ line.

The next one is the $K_{\alpha }$ line and the last one is the ${\text{L}}_{\alpha }$ line. In terms of the wavelength, we can rank the wavelengths from smallest to greatest as, $\lambda \left(K_{\beta }\right)<\lambda \left(K_{\alpha }\right)<\lambda \left(L_{\alpha }\right)$.