Characteristic X-rays

Atomic nuclei emit characteristic X-rays by absorbing energy that excites inner electron shells and allows them to return to lower orbitals. Electrons in such an atom's innermost orbital (e.g., K) may be ejected, leading to the fall-down of higher energy levels' electrons attracting K-alpha x-ray emission attendant on every atom type. Consequently producing distinct spectra corresponding to elements in terms of unique wavelengths characteristic of the radiation with respect to different elements.

Characteristic X-rays play an important role in understanding the structure of atoms and their nuclei. When high-energy electrons crash into atoms, they might knock inner-shell electrons out of place resulting in gaps or vacancies. The energy that was taken up by an electron at a high level escapes it as x-rays named after the object’s element. Over the last ten years of the JEE MAIN exam (from 2013 to 2023), a total of one question has been asked on this concept.

Characteristic X-Rays

Few of the fast-moving electrons have high velocity, penetrate the surface atoms of the target material, and knock out the tightly bound electrons, even from the innermost shells of the atom. Now when the electron is knocked out, a vacancy is created at that place. To fill this vacancy electrons from higher shells jump to fill the created vacancies, we know that when an electron jumps from a higher energy orbit E₁ to a lower energy orbit E₂, it radiates energy (E₁−E₂). Thus this energy difference is radiated in the form of X-rays of very small but definite wavelength which depends upon the target material. The X-ray spectrum consists of sharp lines and is called the characteristic X-ray spectrum. These X-rays are called characteristic X-rays because they are characteristic of the element used as the target anode. Characteristic X-rays have a line spectral distribution, unlike continuous X-rays. The wavelength spectrum of the X-frequencies corresponding to these lines is the characteristic of the material or the target, i.e., anode material.

When the atoms of the target material are bombarded with high-energy electrons (or hard X-rays), which possess enough energy to penetrate into the atom, they knock out the electron of the inner shell (say K shell, n=1 ). When an electron is missing in the K shell, an electron from the next upper shell makes a quantum jump to fill the vacancy in the K shell. In the transition process, the electron radiates
energy whose frequency lies in the X-ray region. The frequency of emitted radiation (i.e., of the photon) is given by -

$v=R Z_e^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Another vacancy is now created in the L shell which is again filled up by another electron jump from one of the upper shell M which results in the emission of another photon, but of different X-ray frequency. This transition continues till outer shells are reached, thus resulting in the emission of a series of spectral lines. The transitions of electrons from various outer shells to the innermost K shell produce a group of X-ray lines called as K -series. These radiations are the most energetic and most penetrating. K-series is further divided into $K_\alpha, K_\beta, K_\gamma, \ldots$ depending upon the outer shell from which the transition is made (see figure).

An incident electron is also known as a projectile electron

Emitted electron is known as photo-electron / orbital electron

Similarly, the rest of the series can be shown as below

Now notice the graph shown below and the sharp peaks obtained in the graph are known as characteristic X-rays because they are characteristic of the target material. The characteristic wavelengths of the material having atomic number Z are called characteristic X-rays and the spectrum obtained is called a characteristic spectrum. If a target material of atomic number $Z^{\prime}$ is used, then peaks are shifted.

The characteristic wavelengths of the material having atomic number Z are called characteristic X-rays and the spectrum
obtained is called a characteristic spectrum. If a target material of atomic number $Z^{\prime}$ is used, then peaks are shifted as shown below

X-ray Absorption

The intensity of X-rays at any point may be defined as the energy falling per second per unit area held perpendicular to the direction of energy flow. Let $I_0$ be the intensity of the incident beam and I will be the intensity of the beam after penetrating a thickness x of a material, then $I=I_0 e^{-\mu x}$, where $\mu$ is the coefficient of absorption or absorption coefficient of the material. The absorption coefficient depends upon the wavelength of X-rays, the density of the material, and the atomic number of the material. The elements of high atomic mass and high density absorb X-rays to a higher degree.

Recommended topic video

Solved Examples Based on Characteristic X-Rays

Example 1: Which of the following is not the application of X-ray?

1) Diffraction

2) To analyse the composition of material

3) Radio communication

4) Radiotherapy

Solution:

Application of X-ray

• X-Ray diffraction
• To analyse the composition of material
• Radiotherapy
• Medicine & surgery

Hence, the answer is the option (3).

Example 2: The wavelength of characteristics $X$-ray $\kappa_\alpha$ line emitted by hydrogen like element is $0.32 A^0$. The wavelength of ${ }^{\kappa_\beta}$ line emitted by the same element is:

1) $0.21 A^{\circ}$
2) $0.27 A^{\circ}$
3) $0.36 A^{\circ}$
4) $0.41 A^{\circ}$

Solution:

Characteristics x-ray

$\nu(\text { frequency })_{=} R z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

wherein

Characteristics X-ray can be defined by knocking out of electron from the inner orbit of an atom

$\begin{aligned} & \kappa_\alpha: \quad \frac{1}{\lambda_\alpha}=R z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R z^2\left(\frac{3}{4}\right) \\ & \kappa_\beta: \quad \frac{1}{\lambda_\beta}=R z^2\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R z^2\left(\frac{8}{9}\right) \\ & \frac{\lambda_\beta}{\lambda_\alpha}=\frac{27}{32} \\ & \lambda_\beta=0.27 A^{\circ}\end{aligned}$

Hence, the answer is the option (2)

Example 3: The wavelengths of Ka x-rays of two metals ‘A’ and ‘B’ are $\frac{4}{1875 P^{\prime}}$ and $\frac{4}{675 R}$ respectively, where ‘R’ is Rydberg's constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is

1) 3

2) 6

3) 5

4) 4

Solution:

According to Moseley's equation

$\frac{1}{\lambda}=R(z-1)^2\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$

For $k_\alpha$ particle; $n_1=2, n_2=1$

$
\frac{1}{\lambda}=R(z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]
$

For metal A;

$
\begin{aligned}
& \frac{\text { }}{} \frac{1875 R}{4}=R\left(Z_1-1\right)^2\left(\frac{3}{4}\right) \\
& \Rightarrow z_1=26
\end{aligned}
$

For metal B;

$
\frac{675 R}{4}=R\left(Z_2-1\right)^2\left(\frac{3}{4}\right) \Rightarrow z_2=31
$

Therefore, 4 elements lie between A and B, i.e. with Z= 27, 28,29, 30

Hence, the answer is the option (4).

Example 4: The intensity of gamma radiation from a given source is $I$. On passing through 36 mm of lead it is reduced to $I / 8$. The thickness (in mm ) of lead which will reduce the intensity to $I / 2$ will be
1) 12
2) 18
3) 6
4) 9

Solution:

Absorption of x-ray

$
I=I_0 e^{-\mu x}
$

wherein
$x=$ thickness of the material

$
\begin{aligned}
& \mu=\text { absorption coefficient } \\
& I=I_0 e^{-\mu x} \\
& \frac{I}{8}=I_0 e^{-\mu(36 m m)} \\
& e^{-36 \mu}=\frac{1}{8} \\
& \frac{I_0}{2}=I_0 e^{-\mu x} \\
& e^{-\mu x}=\frac{1}{2}=\left(\frac{1}{8}\right)^{1 / 3}
\end{aligned}
$

From equation (1)

$
e^{-\mu x}=e^{-12 \mu}
$
Hence, x = 12 mm

Example 5: The $\mathrm{K}_\alpha \mathrm{X}$ - ray of molybdenum has wavelength 0.071 nm . If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV , the energy of this atom when an L electron is knocked out will be $\qquad$ keV . (Round off to the nearest integer)

$
\left[\mathrm{h}=4.14 \times 10^{-15} \mathrm{eVs}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\right]
$

1) 10
2) 11
3) 16
4) 17

Solution:

$\begin{aligned} & E_1 \text { - Energy when an } \mathrm{L} \text { shell-electron is knocked out } \\ & E_2 \text { - Energy when a } \mathrm{K} \text { shell-electron is knocked out } \\ & E_2=27.5 \mathrm{keV} \\ & K_\alpha \rightarrow \text { the wavelength emitted during knocking out of L-electron } \\ & E_{K_\alpha}=\frac{1240}{\lambda_{k \alpha}}=\frac{1240(\mathrm{ev})}{0.071} \\ & E_{k \alpha}=\frac{1240}{71 \times 10^{-3}}=\frac{1240 \times 10^3}{71}(\mathrm{ev}) \\ & E_2=E_1+E_{k_\alpha} \\ & 27.5 \mathrm{eV}=E_1+17.5 \mathrm{keV} \\ & E_1=10 \mathrm{eV}\end{aligned}$

Summary

Each element has particular and unique X-rays that are formed through eliminating an electron from a nucleus’s inner shell generating a hole. Then, an electron from the outer shell fills up this space releasing energy in terms of x-rays. Unlike any other element, the energy emitted during the process of transition is different, solely depending on the characteristics possessed by it concerning various energy backs. These X-rays happen to be quite useful for purposes of material analysis as well as in medical imaging.