X-Rays Questions in English

(B) The shortest wavelength (cut-off wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Substituting the values: $h = 6.62 \times 10^{-34} \text{ J-s}$, $c = 3 \times 10^8 \text{ m/s}$, $e = 1.6 \times 10^{-19} \text{ C}$, and $V = 25,000 \text{ V}$.
$\lambda_{\min} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 25,000}$.
$\lambda_{\min} = \frac{19.86 \times 10^{-26}}{40,000 \times 10^{-19}} = \frac{19.86 \times 10^{-26}}{4 \times 10^{-15}} = 4.965 \times 10^{-11} \text{ m}$.
Converting to $\mathring{A}$: $4.965 \times 10^{-11} \text{ m} = 0.4965 \times 10^{-10} \text{ m} \approx 0.50 \mathring{A}$.

(C) The minimum wavelength of $X$-rays produced is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Substituting the values of $h$,$c$,and $e$,we get the relation: $\lambda_{\min} (\text{in } \mathring{A}) = \frac{12400}{V (\text{in volts})}$.
Given $\lambda_{\min} = 0.1 \mathring{A}$.
Therefore,$0.1 = \frac{12400}{V}$.
$V = \frac{12400}{0.1} = 124000 \text{ V}$.
Converting to kilovolts $(kV)$,$V = 124 \text{ kV}$.

(A) Moseley's law states that the square root of the frequency of characteristic $X$-rays is directly proportional to the atomic number $(Z)$ of the target element.
Mathematically,this is expressed as $\sqrt{f} = a(Z - b)$,where $a$ and $b$ are constants.
Squaring both sides,we get $f = a^2(Z - b)^2$.
Since $a^2$ is also a constant,the law is represented as $f = a(Z - b)^2$.

(B) The penetrating power of $X-$rays is directly related to their energy.
According to the relation $E = eV$,where $V$ is the accelerating potential difference,the energy of the $X-$ray photons increases as the potential difference increases.
Since higher energy photons have a shorter wavelength and higher frequency,they possess greater penetrating power.
Therefore,the penetrating power depends on the applied potential difference.

(D) Characteristic $X-$ rays are produced when a high-energy incident electron knocks out an inner-shell electron (e.g.,from the $K$ shell) of the target atom.
This creates a vacancy in the inner shell,making the atom unstable.
An electron from a higher energy level (e.g.,the $L$ shell) then drops down to fill this vacancy.
The energy difference between these two shells is released in the form of a photon,which constitutes the characteristic $X-$ ray.
Therefore,the energy of the photon comes from the electronic transition within the target atom.

(B) The minimum wavelength (cut-off wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Substituting the given values:
$h = 6.6 \times 10^{-34} \ J \cdot s$
$c = 3 \times 10^8 \ m/s$
$e = 1.6 \times 10^{-19} \ C$
$V = 30 \times 10^3 \ V$
$\lambda_{\min} = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{(1.6 \times 10^{-19}) \times (30 \times 10^3)}$
$\lambda_{\min} = \frac{19.8 \times 10^{-26}}{48 \times 10^{-16}} = 0.4125 \times 10^{-10} \ m = 0.4125 \ \mathring{A}$.
Rounding to the nearest option,we get $0.4 \ \mathring{A}$.

(D) The most energetic $X$-ray corresponds to the minimum wavelength $(\lambda_{\min})$ produced when the entire kinetic energy of the electron is converted into a single photon.
Using the Duane-Hunt law,the minimum wavelength is given by the formula:
$\lambda_{\min} = \frac{hc}{E} = \frac{12400 \text{ eV} \cdot \mathring{A}}{E \text{ (in eV)}}$
Given the energy $E = 100 \text{ keV} = 100,000 \text{ eV}$.
Substituting the values:
$\lambda_{\min} = \frac{12400}{100000} \mathring{A} = 0.124 \mathring{A}$
Thus,the correct option is $D$.

(D) The shortest wavelength (cut-off wavelength) of $X$-rays produced is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ \text{eV} \cdot \mathring{A}}{V \text{ (in volts)}}$.
Given the potential difference $V = 50000 \ V$.
Substituting the values: $\lambda_{\min} = \frac{12400}{50000} \ \mathring{A} = 0.248 \ \mathring{A} \approx 0.0248 \ nm$.
Rounding to the nearest provided option,we get $0.025 \ nm$.

(C) Hard $X$-rays are defined as $X$-rays with higher energy and shorter wavelengths compared to soft $X$-rays.
Since the energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$,a higher energy $E$ implies a higher frequency $\nu$ and a shorter wavelength $\lambda$.
Therefore,for harder $X$-rays,the frequency is higher and the wavelength is shorter.
Thus,option $C$ is correct.

(A) Cathode rays consist of high-speed electrons. When these high-energy electrons strike a metal target with a high melting point (such as tungsten or molybdenum),they are suddenly decelerated by the strong electric fields of the target nuclei. This rapid deceleration causes the emission of high-energy electromagnetic radiation known as $X$-rays. This is the fundamental principle behind the operation of an $X$-ray tube.

(A) The penetrating power of $X-$rays is directly proportional to their energy (frequency).
When the potential difference between the anode and cathode is increased,the kinetic energy of the electrons striking the target increases.
This higher kinetic energy results in the production of $X-$rays with higher energy and shorter wavelengths,which possess greater penetrating power.
Therefore,increasing the potential difference between the anode and cathode increases the penetrating power of $X-$rays.

(A) In the characteristic $X$-ray spectrum,the $K$-series corresponds to transitions where the final state is the $K$-shell $(n = 1)$.
The $K_{\alpha}$ line specifically refers to the transition from the nearest shell,which is the $L$-shell $(n = 2)$,to the $K$-shell $(n = 1)$.
Therefore,the transition is from $n = 2$ to $n = 1$.

(B) The continuous $X$-ray spectrum is produced due to the deceleration of electrons hitting the target. The maximum energy of a photon emitted is equal to the kinetic energy of the incident electron,given by $E_{\max} = eV = \frac{hc}{\lambda_{\min}}$.
Thus,the minimum wavelength is $\lambda_{\min} = \frac{hc}{eV}$.
Since electrons can lose any amount of energy less than their total kinetic energy,photons of all wavelengths greater than $\lambda_{\min}$ are emitted.
Therefore,the range of wavelengths for the continuous $X$-ray spectrum is from $\lambda_{\min}$ to $\infty$.

(B) The wavelength range of $X-$rays typically lies between $0.1 \mathring A$ and $100 \mathring A$.
Among the given options,$2 \mathring A$ falls within this range.
Therefore,the correct option is $B$.

(B) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
Since $h$ and $c$ are constants,the energy is inversely proportional to the wavelength: $E \propto \frac{1}{\lambda}$.
Let $E_1$ be the energy of the $X$-ray photon with wavelength $\lambda_1 = 1 \mathring{A}$ and $E_2$ be the energy of the visible light photon with wavelength $\lambda_2 = 5000 \mathring{A}$.
The ratio of the energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Substituting the given values,we get $\frac{E_1}{E_2} = \frac{5000 \mathring{A}}{1 \mathring{A}} = 5000$.
Therefore,the ratio is $5000 : 1$.

(B) Moseley's law states that the frequency $\nu$ of a characteristic $X$-ray spectral line is given by the relation: $\sqrt{\nu} = a(Z - b)$.
Here,$Z$ is the atomic number of the element,and $a$ and $b$ are constants that depend on the specific type of spectral line (e.g.,$K$,$L$,etc.).
Squaring both sides of the equation,we get: $\nu = a^2(Z - b)^2$.
Since $b$ is a screening constant and is typically small compared to $Z$ for high-$Z$ elements,the frequency $\nu$ is directly proportional to the square of the atomic number $(Z^2)$.
Therefore,the frequency varies as the square of the atomic number of the element.

(A) The structural analysis of crystals relies on the phenomenon of diffraction. For diffraction to occur, the wavelength of the incident radiation must be comparable to the spacing between the scattering centers (atoms) in the crystal lattice. The interatomic spacing in crystals is typically in the range of $0.1 \, nm$ to $0.5 \, nm$ ($1 \, \text{AA}$ to $5 \, \text{AA}$). $X-$ rays possess wavelengths in this same range, making them ideal for probing the atomic structure of crystals through $X-$ ray diffraction $(XRD)$. Therefore, option $A$ is correct.

(B) $X$-rays and $\gamma$-rays are both electromagnetic radiations.
The fundamental difference lies in their origin.
$X$-rays are produced due to electronic transitions in the inner shells of an atom.
$\gamma$-rays are produced due to nuclear transitions within the nucleus of an atom.
Therefore,the essential distinction is that $\gamma$-rays emanate from the nucleus while $X$-rays emanate from the outer part of the atom.

(D) The energy of an electron accelerated through a potential difference $V$ is given by $E = eV$.
When this electron strikes the target,it produces $X$-rays. The maximum energy of the emitted $X$-ray photon corresponds to the total kinetic energy of the electron,which is $E_{max} = h\nu_{max} = \frac{hc}{\lambda_{min}}$.
Equating the two,we get $eV = \frac{hc}{\lambda_{min}}$.
Rearranging for the minimum wavelength,we find $\lambda_{min} = \frac{hc}{eV}$.
Since $h$,$c$,and $e$ are constants,it follows that $\lambda_{min} \propto \frac{1}{V}$.

(D) The hardness of $X-$rays refers to their penetrating power,which is directly related to their energy.
According to the Duane-Hunt law,the maximum frequency $f_{max}$ of $X-$rays produced is given by $hf_{max} = eV$,where $e$ is the elementary charge and $V$ is the accelerating potential difference between the cathode and the target.
Since $E = hf$,higher potential difference $V$ results in higher energy $X-$rays,which are termed 'hard' $X-$rays.
Therefore,the potential difference between the cathode and the target determines the hardness of the $X-$rays.

(D) The intensity of $X-$rays is directly proportional to the number of electrons striking the anode per unit time. When the number of electrons striking the anode increases,the number of $X-$ray photons produced also increases,which leads to an increase in the intensity of the $X-$rays. Other parameters like frequency,wavelength,and penetration power depend on the accelerating voltage (potential difference) applied across the tube,not on the number of electrons.

(A) The minimum wavelength of emitted $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Substituting the values $h = 6.625 \times 10^{-34} \text{ J-s}$, $c = 3 \times 10^8 \text{ m/s}$, and $e = 1.6 \times 10^{-19} \text{ C}$, we get:
$\lambda_{\min} = \frac{12400}{V} \text{ Å}$ (where $V$ is in volts).
Given $\lambda_{\min} = 1 \text{ Å}$, we have $1 = \frac{12400}{V}$.
Therefore, $V = 12400 \text{ V} = 12.4 \text{ kV}$.
Comparing this with the given options, the closest value is $12.42 \text{ kV}$.

(C) For diffraction to occur,the wavelength of the incident wave must be comparable to the spacing between the lines of the grating.
$X$-rays have very short wavelengths (typically $0.01 \ nm$ to $10 \ nm$).
An ordinary optical grating has a spacing of the order of $10^3 \ nm$ to $10^4 \ nm$.
Since the wavelength of $X$-rays is much smaller than the grating spacing,they cannot be diffracted by an ordinary grating.
Therefore,the correct option is $C$.

(C) Statement $A$ is true: The characteristic $X-$ray spectrum is produced due to electronic transitions between energy levels of the target atoms. Since these energy levels $(E_K, E_L, \dots)$ are specific to the atomic number $(Z)$ of the element,the wavelengths $\lambda = \frac{hc}{E_K - E_L}$ depend on the nature of the target material.
Statement $B$ is true: The short wavelength limit (cut-off wavelength) of the continuous $X-$ray spectrum is given by $\lambda_{min} = \frac{hc}{eV}$,where $V$ is the accelerating potential difference. Thus,$\lambda_{min} \propto \frac{1}{V}$,meaning it varies inversely with the potential difference applied to the $X-$ray tube.
Therefore,both statements are true.

(C) The energy $E$ of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given:
$h = 6.6 \times 10^{-34} \ J \cdot s$
$c = 3 \times 10^8 \ m/s$
$\lambda = 1.65 \mathring{A} = 1.65 \times 10^{-10} \ m$
Substituting the values:
$E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.65 \times 10^{-10}} \ J$
$E = \frac{19.8 \times 10^{-26}}{1.65 \times 10^{-10}} \ J = 12 \times 10^{-16} \ J$
To convert this energy into $eV$,divide by $1.6 \times 10^{-19} \ J/eV$:
$E(eV) = \frac{12 \times 10^{-16}}{1.6 \times 10^{-19}} = 7.5 \times 10^3 \ eV = 7.5 \ keV$.
Alternatively,using the shortcut $E(eV) = \frac{12400}{\lambda(\mathring{A})}$,we get $E = \frac{12400}{1.65} \approx 7515 \ eV \approx 7.5 \ keV$.

(B) Bragg's law describes the condition for constructive interference of $X$-rays scattered by the atomic planes of a crystal.
When an $X$-ray beam is incident on a crystal at an angle $\theta$ with respect to the atomic planes,the path difference between the rays reflected from successive planes separated by a distance $d$ is $2d \sin \theta$.
For constructive interference,this path difference must be an integer multiple of the wavelength $\lambda$.
Thus,the condition is $2d \sin \theta = n\lambda$,where $n$ is an integer representing the order of reflection.

(A) The minimum wavelength (cutoff wavelength) of $X-$rays produced by an electron accelerated through a potential difference $V$ is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ \mathring{A} \cdot V}{V \text{ (in volts)}}$.
Given $V = 40 \ kV = 40,000 \ V$.
Substituting the values: $\lambda_{\min} = \frac{12400}{40000} \ \mathring{A} = 0.31 \ \mathring{A}$.
Converting to meters: $0.31 \times 10^{-10} \ m = 3.1 \times 10^{-11} \ m$.
Given the options provided,the closest value is $3.09 \times 10^{-11} \ m$.

(D) The target material used in an $X$-ray tube must possess two primary characteristics:
$1$. A high atomic number $(Z)$ to increase the efficiency of $X$-ray production (Bremsstrahlung radiation).
$2$. A high melting point to withstand the intense heat generated by the bombardment of high-energy electrons.
Tungsten $(W)$ has a high atomic number $(Z = 74)$ and a very high melting point $(3422 ^\circ C)$, making it the ideal choice for the target material.

(A) The intensity of $X-$rays is directly proportional to the number of electrons striking the target material per unit time.
When high-speed electrons collide with the target,they produce $X-$ray photons.
Therefore,increasing the number of electrons increases the number of $X-$ray photons produced,which corresponds to higher intensity.

(B) The energy $E$ of the bombarding electrons is equal to the maximum energy of the $X$-ray photons produced,which corresponds to the minimum wavelength $\lambda_{min}$.
Using the relation $E = \frac{hc}{\lambda_{min}}$,where $h = 6.626 \times 10^{-34} \; J\cdot s$,$c = 3 \times 10^8 \; m/s$,and $\lambda_{min} = 1 \; \mathring{A} = 10^{-10} \; m$.
In terms of electron-volts $(eV)$,the formula is $E(eV) = \frac{12400}{\lambda(\mathring{A})}$ (using $hc \approx 12400 \; eV\cdot\mathring{A}$).
Substituting the given value: $E = \frac{12400}{1} = 12400 \; eV$.
Given the options provided and the standard approximation $hc \approx 12375 \; eV\cdot\mathring{A}$,we calculate: $E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.602 \times 10^{-19} \times 10^{-10}} \approx 12415 \; eV$.
Based on the provided options,the closest value is $12375 \; eV$.

(C) The energy of the incident electron beam is given by $E = eV = 60000 \ eV$.
The binding energy of the $K$-shell electron is $40000 \ eV$.
Since the energy of the incident electrons $(60000 \ eV)$ is greater than the binding energy of the $K$-shell electrons $(40000 \ eV)$,the incident electrons have sufficient energy to knock out the $K$-shell electrons.
When a $K$-shell electron is ejected,an electron from a higher energy level transitions to the $K$-shell,emitting characteristic $X$-rays of all series $(K, L, M, \dots)$.
Additionally,the deceleration of incident electrons in the target produces continuous $X$-rays (Bremsstrahlung).
Therefore,both continuous and all series of characteristic $X$-rays will be formed.

(C) Characteristic $X$-rays are produced when an electron from a higher energy shell transitions to a vacancy in an inner shell.
For the $K$-series,the vacancy is in the $K$-shell $(n = 1)$.
- The $K_{\alpha}$ $X$-ray is produced by a transition from the $L$-shell $(n = 2)$ to the $K$-shell $(n = 1)$.
- The $K_{\beta}$ $X$-ray is produced by a transition from the $M$-shell $(n = 3)$ to the $K$-shell $(n = 1)$.
- The $K_{\gamma}$ $X$-ray is produced by a transition from the $N$-shell $(n = 4)$ to the $K$-shell $(n = 1)$.
Therefore,for the $K_{\beta}$ $X$-ray,the transition is from $n = 3$ to $n = 1$.

(D) The penetrating power of $X-$rays is determined by their energy,which is directly related to their frequency and inversely related to their wavelength.
Increasing the potential difference between the cathode and anode increases the kinetic energy of the electrons,which in turn increases the energy and penetrating power of the $X-$rays.
However,the current in the filament only controls the number of electrons emitted from the cathode (thermionic emission).
Increasing the filament current increases the intensity (number of photons) of the $X-$rays produced,but it does not change the energy of individual photons.
Therefore,the penetrating power does not depend on the current in the filament.

(A) The current $i$ is defined as the rate of flow of charge,given by the formula $i = \frac{Ne}{t}$,where $N$ is the number of electrons,$e$ is the elementary charge $(1.6 \times 10^{-19} \text{ C})$,and $t$ is the time.
To find the number of electrons striking the target per second,we rearrange the formula to solve for $\frac{N}{t}$:
$\frac{N}{t} = \frac{i}{e}$
Given $i = 3.2 \text{ mA} = 3.2 \times 10^{-3} \text{ A}$ and $e = 1.6 \times 10^{-19} \text{ C}$:
$\frac{N}{t} = \frac{3.2 \times 10^{-3}}{1.6 \times 10^{-19}}$
$\frac{N}{t} = 2 \times 10^{16} \text{ electrons/sec}$.

(D) Characteristic $X$-rays are produced when an electron from a higher energy shell transitions to a vacancy in an inner shell of an atom.
For the $K$-series,the vacancy is in the $n = 1$ shell.
The specific transitions are defined as follows:
$K_{\alpha}: n = 2 \to n = 1$
$K_{\beta}: n = 3 \to n = 1$
$K_{\gamma}: n = 4 \to n = 1$
As shown in the diagram,the $K_{\gamma}$ transition corresponds to an electron moving from the $n = 4$ shell to the $n = 1$ shell.
Therefore,the correct option is $(D)$.

(A) $X-$rays are electromagnetic waves consisting of high-energy photons.
Photons are electrically neutral particles,meaning they carry no net electric charge.
The Lorentz force acting on a particle moving in a magnetic field is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the charge $q = 0$ for $X-$rays,the magnetic force $\vec{F}$ acting on them is zero.
Therefore,$X-$rays do not experience any deflection when passing through a strong uniform magnetic field.

(D) The minimum wavelength $(\lambda_{\min})$ of $X$-rays produced in an $X$-ray tube is given by the Duane-Hunt law,where the entire kinetic energy of the electron is converted into a photon of maximum frequency:
$E = eV = \frac{hc}{\lambda_{\min}}$
Therefore,$\lambda_{\min} = \frac{hc}{eV}$.
Substituting the values of Planck's constant $(h = 6.626 \times 10^{-34} \; J \cdot s)$,speed of light $(c = 3 \times 10^8 \; m/s)$,and charge of an electron $(e = 1.6 \times 10^{-19} \; C)$:
$\lambda_{\min} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times V} \; m$
$\lambda_{\min} \approx \frac{12400}{V} \; \mathring{A}$.

(C) The primary difference between soft and hard $X-$rays lies in their energy and frequency. Hard $X-$rays have higher energy and higher frequency compared to soft $X-$rays. Soft $X-$rays have lower energy and lower frequency,making them less penetrating. Therefore,the correct option is $(c)$.

(B) The penetration power of $X$-rays depends on the density and atomic number of the material through which they pass.
Materials with higher density and higher atomic number absorb $X$-rays more effectively,thereby reducing their penetration distance.
Among the given options,iron has the highest density and atomic number compared to air,wood,and water.
Therefore,$X$-rays will travel the minimum distance in iron.

(A) The minimum wavelength $(\lambda_{\min})$ of $X$-rays produced in an $X$-ray tube is given by the relation: $\lambda_{\min} = \frac{12400}{V} \mathring{A}$ (where $V$ is in volts).
Using the more precise constant $12375$ for the given value:
$\lambda_{\min} = \frac{12375}{V} \mathring{A}$
Given $\lambda_{\min} = 0.4125 \mathring{A}$.
Substituting the value:
$0.4125 = \frac{12375}{V}$
$V = \frac{12375}{0.4125} = 30000 \text{ V} = 30 \text{ kV}$.
Therefore,the accelerating voltage is $30 \text{ kV}$.

(B) Characteristic $X$-rays are produced when an element is bombarded with high-energy particles such as photons,electrons,or ions.
When an incident particle strikes a bound electron in an atom,the bound electron is ejected from an inner shell of the atom.
After the electron is ejected,the atom is left with a vacant energy level,known as a core hole.
Electrons from outer shells then fall into the inner shell,emitting quantized photons with energy equivalent to the difference between the higher and lower energy states.
Since each element has a unique set of energy levels,the transitions from higher to lower energy levels produce $X$-rays with frequencies that are characteristic to each specific element.

(D) $X$-rays are electromagnetic waves that carry both energy and momentum. When they are incident on a metal surface,they interact with the electrons and atoms of the metal.
$1$. Energy Transfer: The $X$-rays transfer their energy to the metal surface through processes like the photoelectric effect or Compton scattering.
$2$. Force and Pressure: Since $X$-rays carry momentum $(p = E/c)$,when they are absorbed or reflected by the metal,they exert a radiation pressure on the surface. This pressure results in a force acting on the metal surface.
Therefore,all the given statements are correct.

(A) The minimum wavelength $(\lambda_{\min})$ of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ \text{eV} \cdot \mathring{A}}{V \text{ (in volts)}}$.
Given the potential difference $V = 40 \ kV = 40,000 \ V$.
Substituting the values: $\lambda_{\min} = \frac{12400}{40000} \ \mathring{A} = 0.31 \ \mathring{A}$.
Thus,the minimum wavelength is $0.31 \ \mathring{A}$.

(D) The minimum wavelength (cut-off wavelength) of $X$-rays produced in a Coolidge tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Here,$h = 6.63 \times 10^{-34} \ J \cdot s$ (Planck's constant),$c = 3 \times 10^8 \ m/s$ (speed of light),$e = 1.6 \times 10^{-19} \ C$ (charge of an electron),and $V = 100 \ kV = 10^5 \ V$.
Substituting the values:
$\lambda_{\min} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 10^5} \ m$.
$\lambda_{\min} = \frac{19.89 \times 10^{-26}}{1.6 \times 10^{-14}} \ m$.
$\lambda_{\min} \approx 12.43 \times 10^{-12} \ m = 0.1243 \ \mathring A$.
Rounding to the nearest provided option,we get $0.123 \ \mathring A$.

(D) The maximum frequency of $X$-rays produced in an $X$-ray tube is determined by the Duane-Hunt law,which states that the maximum frequency $\nu_{\max}$ is given by $\nu_{\max} = \frac{eV}{h}$.
However,the question refers to the characteristic $X$-rays. According to Moseley's law,the frequency $\nu$ of characteristic $X$-rays is given by $\nu = a(Z - b)^2$.
For the $K_{\alpha}$ line,the screening constant $b = 1$. Thus,the frequency is proportional to $(Z - 1)^2$.
Given the options provided,the question specifically targets the relationship defined by Moseley's law for characteristic radiation.

(B) $X-$rays are electromagnetic waves.
All electromagnetic waves travel at the speed of light in a vacuum,which is approximately $3 \times 10^8 \text{ m/s}$.
The operating potential of the $X-$ray tube affects the energy and frequency of the $X-$rays produced,but it does not change their velocity.
Therefore,the velocity of the $X-$rays is $3 \times 10^8 \text{ m/s}$.

(B) The intensity of $X$-rays produced in an $X$-ray tube depends on the number of electrons striking the target per unit time,which is determined by the filament current (thermionic emission).
Changing the accelerating voltage (potential difference) affects the kinetic energy of the electrons and the minimum wavelength (cutoff wavelength) of the $X$-rays,but it does not significantly change the total intensity of the $X$-rays produced.
Therefore,if the voltage is doubled,the intensity of the $X$-rays remains unchanged.

(D) The minimum wavelength (cutoff wavelength) of an $X-ray$ tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Substituting the values $h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $e = 1.6 \times 10^{-19} \ C$,we get the simplified relation: $\lambda_{\min} (\text{in } \mathring{A}) \approx \frac{12400}{V (\text{in volts})}$.
Given $\lambda_{\min} = 2.5 \times 10^{-10} \ m = 2.5 \ \mathring{A}$.
Substituting this into the formula: $2.5 = \frac{12400}{V}$.
$V = \frac{12400}{2.5} = 4960 \ V$.
$V \approx 5000 \ V = 5 \ kV$.

(C) The minimum wavelength (cutoff wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$,where $eV$ is the kinetic energy $(K.E.)$ of the electrons striking the target.
From this relation,we can see that $\lambda_{\min} \propto \frac{1}{K.E.}$.
Therefore,when the kinetic energy $(K.E.)$ of the electrons striking the target is increased,the wavelength $\lambda_{\min}$ decreases.

(C) $X$-rays are electromagnetic radiations of very short wavelength. In a laboratory,they are produced using an $X$-ray tube. In this process,a beam of high-energy electrons is accelerated by a high potential difference and made to strike a target made of a heavy metal with a high atomic number (such as Tungsten or Molybdenum). When these fast-moving electrons strike the target,they are suddenly decelerated,and their kinetic energy is converted into $X$-ray photons. Therefore,the correct option is $C$.