X-Rays Questions in English

(B) Moseley's law states that the square root of the frequency $(v)$ of a characteristic $X$-ray spectral line is directly proportional to the atomic number $(Z)$ of the target element.
Mathematically,it is expressed as: $\sqrt{v} = a(Z - b)$.
Here,$a$ and $b$ are constants depending on the specific spectral line (such as $K_\alpha$).
Rearranging this equation,we get: $\frac{\sqrt{v}}{(Z - b)} = a$.
Comparing this with the given options,where $A$ and $B$ are constants,the relation is $\frac{\sqrt{v}}{(Z - A)} = B$.

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given, $\lambda = 45 \times 10^{-2} \text{ Å} = 45 \times 10^{-12} \text{ m}$.
Substituting the values: $E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{45 \times 10^{-12}} \text{ J}$.
$E = \frac{19.86 \times 10^{-26}}{45 \times 10^{-12}} \text{ J} = 0.4413 \times 10^{-14} \text{ J}$.
To convert energy into $eV$, divide by the charge of an electron $(1.6 \times 10^{-19} \text{ C})$:
$E_{eV} = \frac{0.4413 \times 10^{-14}}{1.6 \times 10^{-19}} \text{ eV} \approx 0.2758 \times 10^5 \text{ eV} = 27,580 \text{ eV}$.
Rounding to the nearest given option, the value is $27,500 \text{ eV}$.

(A) The minimum wavelength $(\lambda_{min})$ of $X$-rays produced by electrons accelerated through a potential difference $V$ is given by the Duane-Hunt law: $\lambda_{min} = \frac{hc}{eV}$.
Given $V = 20 \ kV = 20 \times 10^3 \ V$.
Using the values $h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $e = 1.6 \times 10^{-19} \ C$:
$\lambda_{min} = \frac{12400 \ Å \cdot V}{V \text{ (in volts)}}$.
$\lambda_{min} = \frac{12400}{20000} \ Å = 0.62 \ Å = 0.062 \ nm$.

(A) The electromagnetic spectrum classifies $X$-rays as high-energy radiation with wavelengths typically ranging from $10^{-8} \,m$ to $10^{-12} \,m$.
Among the given options, $10^{-10} \,m$ falls within this range, making it a typical wavelength for $X$-rays.

(A) Given: Atomic number $Z=31$ and constant $a=5 \times 10^7 \text{ Hz}^{1/2}$.
According to Moseley's Law for the $K_{\alpha}$ line:
$\sqrt{\nu} = a(Z-1)$
Squaring both sides:
$\nu = a^2(Z-1)^2$
Substitute the values:
$\nu = (5 \times 10^7)^2 \times (31-1)^2$
$\nu = 25 \times 10^{14} \times 30^2$
$\nu = 25 \times 10^{14} \times 900 = 2.25 \times 10^{18} \text{ Hz}$
Now, using the relation $\lambda = \frac{c}{\nu}$ where $c = 3 \times 10^8 \text{ m/s}$:
$\lambda = \frac{3 \times 10^8}{2.25 \times 10^{18}}$
$\lambda = 1.33 \times 10^{-10} \text{ m}$
Since $1 \text{ Å} = 10^{-10} \text{ m}$, we get:
$\lambda = 1.33 \text{ Å}$.

(B) The Compton shift formula is given by $\Delta \lambda = \lambda^{\prime} - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.
Given $\lambda = 0.140 \,nm = 0.140 \times 10^{-9} \,m$ and $\phi = 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the shift is $\Delta \lambda = \frac{h}{m_e c}$.
The Compton wavelength is $\frac{h}{m_e c} \approx 2.426 \times 10^{-12} \,m = 0.002426 \,nm$.
Thus,$\lambda^{\prime} = \lambda + \Delta \lambda = 0.140 \,nm + 0.002426 \,nm = 0.142426 \,nm$.
Rounding to three decimal places,we get $\lambda^{\prime} \approx 0.142 \,nm$.

(B) Moseley's law states that the square root of the frequency $(v)$ of a characteristic $X$-ray spectral line is directly proportional to the atomic number $(Z)$ of the element.
Mathematically,this is expressed as: $\sqrt{v} = a(Z - b)$,where $a$ and $b$ are constants.
In the given options,if we rearrange the relation $\sqrt{v} = B(Z-A)$,we get $\frac{\sqrt{v}}{(Z-A)} = B$.
Therefore,the correct relation is $\frac{\sqrt{v}}{(Z-A)} = B$.

(D) The wavelength of the $K_\alpha$ line,denoted as $\lambda_{K_\alpha}$,is constant and independent of the operating voltage $V$.
The minimum wavelength of the continuous $X$-ray spectrum is given by $\lambda_{min} = \frac{hc}{eV}$.
Given $\Delta \lambda = \lambda_{min} - \lambda_{K_\alpha} = \frac{hc}{eV} - \lambda_{K_\alpha}$.
When the voltage is changed to $V^{\prime} = V/3$,the new minimum wavelength is $\lambda_{min}^{\prime} = \frac{hc}{e(V/3)} = 3 \frac{hc}{eV} = 3 \lambda_{min}$.
The new difference is $\Delta \lambda^{\prime} = \lambda_{min}^{\prime} - \lambda_{K_\alpha} = 3 \lambda_{min} - \lambda_{K_\alpha}$.
Since $\lambda_{min} = \Delta \lambda + \lambda_{K_\alpha}$,we substitute this into the expression for $\Delta \lambda^{\prime}$:
$\Delta \lambda^{\prime} = 3(\Delta \lambda + \lambda_{K_\alpha}) - \lambda_{K_\alpha} = 3 \Delta \lambda + 3 \lambda_{K_\alpha} - \lambda_{K_\alpha} = 3 \Delta \lambda + 2 \lambda_{K_\alpha}$.
Since $\lambda_{K_\alpha} > 0$,it follows that $\Delta \lambda^{\prime} > 3 \Delta \lambda$. Therefore,the correct option is $D$.

(D) According to Moseley's law for characteristic $X$-rays,the frequency $v$ of the emitted $X$-rays is related to the atomic number $Z$ as $v \propto (Z - b)^2$,where $b$ is a screening constant. For the $K_{\alpha}$ line,$b = 1$.
Assuming the same series for both elements,we have $v \propto Z^2$.
Given the ratio of atomic numbers $Z_{A} : Z_{B} = 1 : 2$.
Therefore,the ratio of frequencies is:
$\frac{v_{A}}{v_{B}} = \left( \frac{Z_{A}}{Z_{B}} \right)^2$
$\frac{v_{A}}{v_{B}} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$
Thus,$v_{A} : v_{B} = 1 : 4$.

(D) When $X$-rays are incident on parallel atomic planes with spacing $d$ at a glancing angle $\theta$,the path difference between the rays reflected from successive planes is given by $\Delta x = 2 d \sin \theta$.
For constructive interference (maxima),the path difference must be an integer multiple of the wavelength $\lambda$.
Therefore,the condition for maxima is $2 d \sin \theta = n \lambda$,where $n = 1, 2, 3, \dots$ is the order of the interference fringe. This is known as Bragg's Law.