The intensity absorbed in a material of depth $d$ with absorption coefficient $\mu $, when ${I_0}$ is the incident intensity of X-rays, is:
A. $\left( {1 - {e^{ - \mu d}}} \right)$
B. ${I_0}\left( {1 - {e^{ - \mu d}}} \right)$
C. $\mu {I_0}\left( {1 - {e^{ - \mu d}}} \right)$
D. $\mu d$
Answer
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Hint:The X-rays belong to the extreme end of the electromagnetic spectrum with frequencies beyond the visible spectrum and the ultraviolet radiation. The reason for getting absorbed by molecules is because its wavelength is similar to the bond distance, and hence, X-ray gets scattered by molecules in the matter.
Complete step-by-step answer:
An X-ray is a penetrating form of high-energy electromagnetic radiation. The X-rays have a wavelength ranging from 10 picometres to 10 nanometres, corresponding to frequencies in the range $3 \times {10^{16}}Hz - 3 \times {10^{19}}Hz$ and energies in the range 124 eV to 124 keV. X-ray wavelengths are shorter than those of UV rays and typically longer than those of gamma rays.
The wavelength of the X-ray being in nanometres, is equivalent to the size of the molecules. Hence, this serves as a necessary condition for scattering of X-ray when passed through a substance. Scattering involves absorption of the X-ray radiation and re-emitting in different directions with different intensities.
Consider a material of depth d. Let ${I_0}$ be the intensity of the incident X-rays and $\mu $ be the absorption coefficient of the material. Then, the intensity of the X-rays that is reduced after travelling through the distance d in the material is given by,
$I = {I_0}{e^{ - \mu d}}$
If $I$ is the intensity that has been reduced after absorption in the material, then the intensity of light absorbed, will be the difference between the incident radiation and the reduced intensity.
Thus,
Absorbed intensity, ${I_a} = {I_0} - I$
Substituting for $I$, we get –
$
{I_a} = {I_0} - I \\
{I_a} = {I_0} - {I_0}{e^{ - \mu d}} \\
\to {I_a} = {I_0}\left( {1 - {e^{ - \mu d}}} \right) \\
$
Hence, the correct option is Option B.
Note: The absorption of X-ray by the molecules is a major principle in the subject of X-ray spectroscopy, which is widely used in analytical chemistry, medical field and in remote sensing, only to name a few.
Complete step-by-step answer:
An X-ray is a penetrating form of high-energy electromagnetic radiation. The X-rays have a wavelength ranging from 10 picometres to 10 nanometres, corresponding to frequencies in the range $3 \times {10^{16}}Hz - 3 \times {10^{19}}Hz$ and energies in the range 124 eV to 124 keV. X-ray wavelengths are shorter than those of UV rays and typically longer than those of gamma rays.
The wavelength of the X-ray being in nanometres, is equivalent to the size of the molecules. Hence, this serves as a necessary condition for scattering of X-ray when passed through a substance. Scattering involves absorption of the X-ray radiation and re-emitting in different directions with different intensities.
Consider a material of depth d. Let ${I_0}$ be the intensity of the incident X-rays and $\mu $ be the absorption coefficient of the material. Then, the intensity of the X-rays that is reduced after travelling through the distance d in the material is given by,
$I = {I_0}{e^{ - \mu d}}$
If $I$ is the intensity that has been reduced after absorption in the material, then the intensity of light absorbed, will be the difference between the incident radiation and the reduced intensity.
Thus,
Absorbed intensity, ${I_a} = {I_0} - I$
Substituting for $I$, we get –
$
{I_a} = {I_0} - I \\
{I_a} = {I_0} - {I_0}{e^{ - \mu d}} \\
\to {I_a} = {I_0}\left( {1 - {e^{ - \mu d}}} \right) \\
$
Hence, the correct option is Option B.
Note: The absorption of X-ray by the molecules is a major principle in the subject of X-ray spectroscopy, which is widely used in analytical chemistry, medical field and in remote sensing, only to name a few.
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