(B) When an $X$-ray beam passes through a material of thickness $dx$,the decrease in intensity $dI$ is proportional to the intensity $I$ and the thick

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(B) When an $X$-ray beam passes through a material of thickness $dx$,the decrease in intensity $dI$ is proportional to the intensity $I$ and the thickness $dx$.This is given by the differential equation: $-dI = \mu I dx$.Rearranging the terms,we get: $\frac{dI}{I} = -\mu dx$.Integrating both sides from $x = 0$ (where $I = I_0$) to $x = d$ (where $I = I$):$\int_{I_0}^{I} \frac{dI}{I} = -\int_{0}^{d} \mu dx$.This results in: $\ln(\frac{I}{I_0}) = -\mu d$.Taking the exponential of both sides,we get: $I = I_0e^{-\mu d}$.

Class 12 Physics Atoms X-Rays

Medium

$X$-ray beam of intensity $I_0$ passes through an absorption plate of thickness $d$. If the absorption coefficient of the material of the plate is $\mu$,the correct statement regarding the transmitted intensity $I$ of the $X$-ray is:

A
$I = I_0(1 - e^{-\mu d})$
B
$I = I_0e^{-\mu d}$
C
$I = I_0(1 - e^{-\mu /d})$
D
$I = I_0e^{-\mu /d}$

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$X$-rays of wavelength $\lambda$ are reflected from parallel planes of atoms in a crystal with spacing $d$ between two planes, as shown in the figure. If the two reflected beams interfere constructively, then the condition for maxima will be (where $n$ is the order of the interference fringe):

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$X$-ray beam of intensity $I_0$ passes through an absorption plate of thickness $d$. If the absorption coefficient of the material of the plate is $\mu$,the correct statement regarding the transmitted intensity $I$ of the $X$-ray is:

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$X$-rays of wavelength $0.1 \ \mathring A$ are allowed to fall on a metal and get scattered. The wavelength of the scattered radiation is $0.111 \ \mathring A$. If $h = 6.624 \times 10^{-34} \ J \cdot s$ and $m_0 = 9 \times 10^{-31} \ kg$,then the direction of the scattered photons will be:

The primary difference between $X$-rays and $\gamma$-rays is that:

An $X$-ray tube produces a continuous spectrum of radiation with its shortest wavelength of $45 \times 10^{-2} \text{ Å}$. The maximum energy of a photon in the radiation in $eV$ is (in $27,500$). $(h = 6.62 \times 10^{-34} \text{ J-s}, c = 3 \times 10^8 \text{ m/s})$.

$X$-rays of wavelength $\lambda$ are reflected from parallel planes of atoms in a crystal with spacing $d$ between two planes, as shown in the figure. If the two reflected beams interfere constructively, then the condition for maxima will be (where $n$ is the order of the interference fringe):

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