# X ray discrete and continous spectrum

1) How many types of X ray spectrum exist?. How are the discrete lines called and which are the corresponding transitions?

2) Using the Moseley law find: the wavelength of the $K_{\alpha}$ line for Al (Z=13) and for Co (Z =27);

The Rydberg constant $R$ in known and the screening constant $\sigma=1$.

1) There exist 2 types of X ray spectrum: continuous and characteristic.

The continuous X ray spectrum appears when an electron previously accelerated by a

potential difference $U$ is deflected by the target atoms. Because this deflection corresponds to a loose in energy of the incoming electrons, this energy is released as X photons. Thus the maximum X rays frequency possible emitted is equal to the maximum energy of the incoming electrons:

$E =e U =h*F =h*c/\lambda$

The second type of X ray spectrum arises when an incoming electron has enough energy to ionize an atom electron from an inner shell. The other atom electrons will rearrange to fill the missing space and a set of X ray lines will be emitted corresponding to these electron transitions form outer shells to inner shells.

The Lines of characteristic spectrum are denoted by K -transitions from outer shells to n=1 shell, L -transitions from outer shells to n=2 shell, etc.

Inside the same group of lines, $\alpha$ denotes a transition between two consecutive levels, $\beta$ denotes a transition skipping one level, etc.. For example $K_{\alpha}$ is corresponds to an electron transition between n=2 and n=1. $L_{\beta}$ corresponds to a transitions between n=4 and n=2 levels.

2) The Moseley law relates the wavelength of the line from the characteristic (line) spectrum to the corresponding transitions of electrons between inner shells.

$1/\lambda = R(Z-\sigma)^2\left [(1/n^2)-(1/k^2) \right ]$

The Rydberg constant is $R =1.097*10^7 (1/m)$

Line $K_{\alpha}$ corresponds to transitions k=2 to n=1

For Al and Co the value of the screening constant is $\sigma =1$.

$\lambda = 4/[3R(Z-1)^2]$

Therefore $\lambda_{Al} =844 pm$ and $\lambda_{Co} =180 pm$.