Physics > Advanced Modern Physics > 1.0 X-Rays
Advanced Modern Physics
1.0 X-Rays
2.0 Moseley's Law.
3.0 Nuclear Structure
4.0 Nuclear binding energy
4.1 Binding energy per nucleon
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability
5.0 Radioactivity
6.0 Radioactive decay law
1.1 Properties and Application of X-Rays
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability
- They are electromagnetic radiation and travel with a speed of light.
- They ionize the gas through which they pass(as they knock out electrons of several neutral atoms leaving them positively charged).
- They have high penetrating power and can pass through several materials which are opaque to ordinary light.
- They are not deflected by electric and magnetic fields implying that they don't consist of charged particles.
- They show all properties of waves except refraction.
Application of X-rays
- Diffraction patterns are used in determining the internal structure of crystals.
- Due to high penetration power, it can be used to detect defects in metallic structures of machines, bridges etc.
- It is used in radio therapy to damage the tissues undergoing malignant growth such as cancer and tumors.
Let the accelerating voltage be $V$.
So, electrons traveling with maximum energy produce X-rays with the minimum wavelength known as cut-off wavelength.
$$\begin{equation} \begin{aligned} eV = h{v_{\max }} = \frac{{hc}}{{{\lambda _{\min }}}} \\ {\text{or }}{\lambda _{\min }} = \frac{{hc}}{{eV}} = \frac{{12400}}{{{V_{\left( {{\text{in volts}}} \right)}}}}{\text{ }}\mathop {\text{A}}\limits^0 \\\end{aligned} \end{equation} $$
Clearly, it don't depends upon the material of the target.
The intensity of X-ray decreases as is penetrating more and more into a material.
Let intensity of the incident beam= ${I_0}$
Intensity after penetration of a thickness $x$= $I$
Absorption coefficient of the material= $a$
It depends on the wavelength of the X-ray incident, the density of the material and the atomic number of the material. The elements with high density and atomic mass absorb more.
So, the relation is, $$I = {I_0}{e^{ - ax}}$$
Question 1. An X-ray tube operates at 20 $KV$. A particular electron loses 5% of its kinetic energy to emit an X-ray photon at the first collision. Find the wavelength corresponding to this
photon.
Solution:
Kinetic energy acquired by the electron is, $$K = 20 \times {10^3}\;{\text{eV}}$$
The energy of the photon is, $$ \Rightarrow 0.05 \times 20 \times {10^3}{\text{ eV}} = {10^3}{\text{ eV}}$$
Therefore, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{hc}}{\lambda } = {10^3}{\text{ eV}} \\ \Rightarrow \lambda = \frac{{12400}}{{1000}} = 12.4{\text{ }}\mathop {\text{A}}\limits^0 \\\end{aligned} \end{equation} $$
Question 2. The voltage applied to an X-ray tube being increased $\eta = 1.5$, the short wave limit of an X-ray continuous spectrum shifts by $\Delta \lambda = 26{\text{ pm}}$. Find the initial voltage applied to the tube.
Solution: $${\lambda _{\min }} = \frac{{ch}}{{eV}}$$ Here, $${\lambda _1} = \frac{{ch}}{{e{V_1}}}\;$$ and $${\lambda _2} = \frac{{ch}}{{e{V_2}}}$$ or $$\begin{equation} \begin{aligned} {\lambda _2} - {\lambda _1} = \frac{{ch}}{e}\left[ {\frac{1}{{{V_2}}} - \frac{1}{{{V_1}}}} \right] \\ \Delta \lambda = 26 \times {10^{ - 12}} \\\end{aligned} \end{equation} $$ So, $$\frac{{ch}}{e}\left[ {\frac{2}{{3{V_1}}} - \frac{1}{{{V_1}}}} \right] = 26 \times {10^{ - 12}}$$
