91Ó°ÊÓ

a) The $K_{\mathrm{Î\pm }}$line arises because of the transition between the K-shell (n=1) and the L-shell (n=2) .

b) The$K_{\mathrm{β}}$line arises because of the transition between the K-shell (n=1) and the M-shell (n=3) .

c) The target is molybdenum with atomic number (Z=42).

d) A single wavelength splits into several wavelengths.

A single wavelength splits into several wavelength components as spectral lines considering the transition from the K-shell to the L-shell or the transition from K-shell to the M-shell. Thus, using the Rydberg equation for wavenumber, we can get the number of components for each transition accordingly.

Formula:

The Rydberg formula for a wavenumber of a transition between two states,
$\overline{v}=R_{H}Z\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),Where{R}_H=1.097×{10}^7{m}^{-1}$ …â¶Ä¦..(¾±)

For a transition, the excitation takes place between states$n_1=1$ and $n_2=2$.

Thus, the number of components of the wavelengths can be given using equation (i) as follows:

$$\begin{gathered} \overline{v}=\left(1.097×{10}^7{m}^{-1}\right)\left(42\right)\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \\ =34.555×{10}^7{m}^{-1} \\ ≈3.5×{10}^8{m}^{-1} \end{gathered}$$

Hence, the required value of components is$3.5×{10}^8$ for a single wavelength.

For a$K_{\mathrm{β}}$ transition, the excitation takes place between states$n_1=1$ and $n_2=3$.

Thus, the number of components of the wavelengths can be given using equation (i) as follows:

$$\begin{gathered} \overline{v}=\left(1.097×{10}^7{m}^{-1}\right)\left(42\right)\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \\ =40.955×{10}^7{m}^{-1} \\ ≈4.1×{10}^8{m}^{-1} \end{gathered}$$

Hence, the required value of components is$4.1×{10}^8$ for a single wavelength.