X-Rays Questions in English

(B) When a high-energy electron strikes a heavy metal target (like tungsten),it undergoes rapid deceleration due to the interaction with the strong electric field of the target nuclei. This process results in the emission of electromagnetic radiation in the $X-$ ray region of the spectrum. Therefore,the emitted radiation is $X-$ rays.

(B) The relationship between frequency $\nu$,speed of light $c$,and wavelength $\lambda$ is given by the formula: $\nu = \frac{c}{\lambda}$.
Given: $c = 3 \times 10^8\,m/s$ and $\lambda = 1\,\mathring{A} = 1 \times 10^{-10}\,m$.
Substituting these values into the formula:
$\nu = \frac{3 \times 10^8}{1 \times 10^{-10}} = 3 \times 10^{18}\,Hz$.
Therefore,the correct option is $B$.

(B) According to Moseley's Law,the frequency $(f)$ of the characteristic $X-$rays emitted by an element is related to its atomic number $(Z)$ by the equation: $f = a(Z - b)^2$,where $a$ and $b$ are constants.
For high atomic numbers,$b$ is negligible,so $f \propto Z^2$.
Therefore,the frequency of characteristic $X-$rays varies with the square of the atomic number.

(A) The minimum accelerating voltage $V$ required to produce $X$-rays of a given wavelength $\lambda$ is determined by the Duane-Hunt law,where the kinetic energy of the electron is converted into the energy of the photon: $eV = \frac{hc}{\lambda}$.
Given:
$h = 6.63 \times 10^{-34} \ \text{J}\cdot\text{s}$
$c = 3 \times 10^8 \ \text{m/s}$
$\lambda = 1.50 \ \mathring{A} = 1.50 \times 10^{-10} \ \text{m}$
$e = 1.6 \times 10^{-19} \ \text{C}$
Substituting the values:
$V = \frac{hc}{e\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 1.50 \times 10^{-10}}$
$V = \frac{19.89 \times 10^{-26}}{2.4 \times 10^{-29}}$
$V = 8.2875 \times 10^3 \ \text{V} \approx 8280 \ \text{V}$.

(D) According to Moseley's Law,the frequency $f$ of the characteristic $X$-rays is given by $f = a(Z - b)^2$.
For the $K_{\alpha}$ line,the screening constant $b = 1$,so $f \propto (Z - 1)^2$.
Since the frequency $f$ is inversely proportional to the wavelength $\lambda$ $(f = \frac{c}{\lambda})$,we have $\frac{c}{\lambda} \propto (Z - 1)^2$.
Therefore,$\lambda \propto \frac{1}{(Z - 1)^2}$.

(D) The absorption of $X$-rays depends on the atomic number $(Z)$ and the density of the material. The absorption coefficient increases significantly with a higher atomic number.
Comparing the atomic numbers of the given elements:
- Beryllium $(Be)$: $Z = 4$
- Copper $(Cu)$: $Z = 29$
- Gold $(Au)$: $Z = 79$
- Lead $(Pb)$: $Z = 82$
Since Lead $(Pb)$ has the highest atomic number among the given options,it provides the maximum absorption of $X$-rays. Therefore,the correct option is $D$.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Here,$h = 6.6 \times 10^{-34} \ J \cdot s$ (Planck's constant),
$c = 3 \times 10^8 \ m/s$ (speed of light),
and $\lambda = 1.54 \mathring A = 1.54 \times 10^{-10} \ m$.
Substituting these values into the formula:
$E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.54 \times 10^{-10}} \ J$
$E = \frac{19.8 \times 10^{-26}}{1.54 \times 10^{-10}} \ J$
$E \approx 12.857 \times 10^{-16} \ J \approx 12.9 \times 10^{-16} \ J$.

(C) According to Moseley's law,the frequency of the ${K_{\alpha}}$ $X$-ray line is given by $\nu = cR(Z-1)^2(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3}{4}cR(Z-1)^2$.
Since $\nu = \frac{c}{\lambda}$,we have $\frac{c}{\lambda} \propto (Z-1)^2$,which implies $\lambda \propto \frac{1}{(Z-1)^2}$.
Given for $Z_1 = 43$,the wavelength is $\lambda_1 = \lambda$.
For $Z_2 = 29$,let the wavelength be $\lambda_2$.
Using the ratio: $\frac{\lambda_2}{\lambda_1} = \left(\frac{Z_1 - 1}{Z_2 - 1}\right)^2$.
Substituting the values: $\frac{\lambda_2}{\lambda} = \left(\frac{43 - 1}{29 - 1}\right)^2 = \left(\frac{42}{28}\right)^2$.
Simplifying the fraction: $\frac{42}{28} = \frac{3}{2}$.
Therefore,$\frac{\lambda_2}{\lambda} = (\frac{3}{2})^2 = \frac{9}{4}$.
Thus,$\lambda_2 = \frac{9}{4}\lambda$.

(A) In an $X-$ray spectrum,$K_{\alpha}$ and $K_{\beta}$ lines are part of the characteristic $X-$ray spectrum.
These lines are produced when an electron from a higher energy shell (like $L$ or $M$ shell) transitions to a vacancy in the $K$ shell $(n=1)$.
Specifically,$K_{\alpha}$ corresponds to a transition from the $L$ shell $(n=2)$ to the $K$ shell $(n=1)$,and $K_{\beta}$ corresponds to a transition from the $M$ shell $(n=3)$ to the $K$ shell $(n=1)$.
Since these lines depend on the atomic number of the target material,they are called characteristic $X-$rays.

(A) The Compton shift formula is given by: $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$.
Here,$\lambda = 0.1 \ \mathring A$ and $\lambda' = 0.111 \ \mathring A$.
So,$\Delta \lambda = 0.111 \ \mathring A - 0.1 \ \mathring A = 0.011 \ \mathring A = 0.011 \times 10^{-10} \ m$.
Rearranging the formula to find $\cos \phi$: $\cos \phi = 1 - \frac{\Delta \lambda \cdot m_0 c}{h}$.
Substituting the values: $\cos \phi = 1 - \frac{0.011 \times 10^{-10} \times 9 \times 10^{-31} \times 3 \times 10^8}{6.624 \times 10^{-34}}$.
$\cos \phi = 1 - \frac{0.0297 \times 10^{-33}}{6.624 \times 10^{-34}} = 1 - \frac{0.297}{0.6624} \approx 1 - 0.4484 = 0.5516$.
Wait,recalculating: $\frac{0.011 \times 10^{-10} \times 9 \times 10^{-31} \times 3 \times 10^8}{6.624 \times 10^{-34}} = \frac{0.0297 \times 10^{-33}}{6.624 \times 10^{-34}} = \frac{29.7}{66.24} \approx 0.4484$.
Therefore,$\cos \phi = 1 - 0.4484 = 0.5516$. Given the options,the intended calculation likely used $m_e = 9.1 \times 10^{-31} \ kg$ and resulted in $0.547$. Thus,$\phi = \cos^{-1}(0.547)$.

(D) According to Bragg's law, the condition for diffraction is $2d \sin \theta = n\lambda$, where $d$ is the interplanar spacing, $\theta$ is the glancing angle, $n$ is the order of diffraction, and $\lambda$ is the wavelength of $X$-rays.
To find the maximum wavelength $\lambda_{\max}$ that can be studied, we set the order of diffraction to $n = 1$ and the maximum value of $\sin \theta$ to $1$ (since $\sin \theta \leq 1$).
Thus, $\lambda_{\max} = 2d$.
Given $d = 10^{-7} \text{ cm} = 10^{-7} \times 10^7 \text{ Å} = 10 \text{ Å}$.
Substituting the value of $d$, we get $\lambda_{\max} = 2 \times 10 \text{ Å} = 20 \text{ Å}$.

(D) The total power supplied is $P = VI = (50 \times 10^3 V) \times (20 \times 10^{-3} A) = 1000 W$.
Since $1\%$ of power is converted into $X$-rays,$99\%$ is converted into heat.
Power converted into heat $P_H = 0.99 \times 1000 W = 990 W$.
The rate of rise of temperature is given by $P_H = ms \frac{dT}{dt}$.
Substituting the values: $990 = (1.0 kg) \times (495 J kg^{-1} {}^\circ C^{-1}) \times \frac{dT}{dt}$.
$\frac{dT}{dt} = \frac{990}{495} = 2 {}^\circ C/s$. Thus,option $(c)$ is correct.
The minimum wavelength is given by $\lambda_{min} = \frac{hc}{eV}$.
$\lambda_{min} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 50 \times 10^3} \approx 0.248 \times 10^{-10} m \approx 0.25 \times 10^{-10} m$. Thus,option $(b)$ is correct.
Since the target absorbs a large amount of heat,it must have a high melting point to prevent damage. Thus,option $(a)$ is correct.
Therefore,all statements are correct.

(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: $I = I_0e^{-\mu d}$.

(C) The energy of the emitted $X$-ray photon corresponds to the energy difference between the two levels involved in the transition.
For the $K_{\alpha}$ line,the transition occurs from the $L$ level to the $K$ level.
The energy difference is given by $\Delta E = E_K - E_L = \frac{hc}{\lambda}$.
Substituting the values: $h = 6.6 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 0.021 \times 10^{-9} \ m$.
$\Delta E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{0.021 \times 10^{-9}} \ J$.
To convert this energy into electron-volts $(eV)$,divide by $1.6 \times 10^{-19} \ J/eV$:
$\Delta E = \frac{19.8 \times 10^{-26}}{0.021 \times 10^{-9} \times 1.6 \times 10^{-19}} \ eV \approx 58928 \ eV \approx 59 \ keV$.

(D) The minimum wavelength of the continuous $X$-ray spectrum is given by the formula $\lambda_{\min} = \frac{hc}{E}$.
Using $hc \approx 12400 \ eV \cdot \mathring{A}$ (or $12375 \ eV \cdot \mathring{A}$),we get $\lambda_{\min} = \frac{12375}{80 \times 10^3} \approx 0.155 \ \mathring{A}$.
Since the energy of the incident electrons $(80 \ keV)$ is greater than the ionization energy of the $K$-shell electrons $(72.5 \ keV)$,the incident electrons have sufficient energy to eject $K$-shell electrons from the tungsten atoms.
When an electron is ejected from the $K$-shell,electrons from higher energy levels transition to fill the vacancy,resulting in the emission of characteristic $X$-rays.
Therefore,the emitted $X$-rays consist of both a continuous spectrum (Bremsstrahlung) and the characteristic $X$-ray spectrum of tungsten.

(A) The wavelength of the $L_{\alpha}$ line is given by Moseley's law for $X$-rays:
$\frac{1}{\lambda} = R(Z - b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For the $L_{\alpha}$ line,$n_1 = 2$ and $n_2 = 3$,and the screening constant $b \approx 7.4$.
Thus,$\frac{1}{\lambda} \propto (Z - 7.4)^2$,which implies $\lambda \propto \frac{1}{(Z - 7.4)^2}$.
Given $\lambda_1 = 1.30 \ \mathring{A}$ for $Z_1 = 78$ and $\lambda_2$ for $Z_2 = 42$:
$\frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 7.4)^2}{(Z_2 - 7.4)^2}$
$\lambda_2 = 1.30 \times \left( \frac{78 - 7.4}{42 - 7.4} \right)^2$
$\lambda_2 = 1.30 \times \left( \frac{70.6}{34.6} \right)^2$
$\lambda_2 = 1.30 \times (2.04)^2 \approx 1.30 \times 4.16 = 5.41 \ \mathring{A}$.

(A) According to Moseley's law,the frequency of $K_{\alpha}$ $X$-rays is given by $\nu = c/\lambda = a(Z - b)^2$,where $b = 1$ for $K_{\alpha}$ lines.
Thus,$\frac{1}{\lambda} \propto (Z - 1)^2$,or $\frac{1}{\sqrt{\lambda}} \propto (Z - 1)$.
For the first atom,$Z_1 = 11$ and $\lambda_1 = \lambda$. So,$\frac{1}{\sqrt{\lambda}} = k(11 - 1) = 10k$.
For the second atom,$Z_2 = Z$ and $\lambda_2 = 4\lambda$. So,$\frac{1}{\sqrt{4\lambda}} = k(Z - 1) = \frac{1}{2\sqrt{\lambda}} = k(Z - 1)$.
Dividing the two equations: $\frac{1/\sqrt{\lambda}}{1/(2\sqrt{\lambda})} = \frac{10k}{k(Z - 1)}$.
$2 = \frac{10}{Z - 1}$.
$Z - 1 = 5$,which gives $Z = 6$.

(C) According to the energy level diagram of $X-$ ray spectra,the energy of a photon emitted during an electronic transition is given by $\Delta E = \frac{hc}{\lambda}$,which implies $\lambda \propto \frac{1}{\Delta E}$.
For $X-$ ray transitions:
$1$. The $K_\beta$ line corresponds to a transition from the $M$ shell to the $K$ shell.
$2$. The $K_\alpha$ line corresponds to a transition from the $L$ shell to the $K$ shell.
$3$. The $L_\alpha$ line corresponds to a transition from the $M$ shell to the $L$ shell.
From the energy conservation principle,the energy of the $K_\beta$ transition is the sum of the energies of the $K_\alpha$ and $L_\alpha$ transitions:
$(\Delta E)_{K_\beta} = (\Delta E)_{K_\alpha} + (\Delta E)_{L_\alpha}$
Substituting $\Delta E = \frac{hc}{\lambda}$ into the equation:
$\frac{hc}{\lambda_\beta} = \frac{hc}{\lambda_\alpha} + \frac{hc}{\lambda'_\alpha}$
Dividing both sides by $hc$,we get:
$\frac{1}{\lambda_\beta} = \frac{1}{\lambda_\alpha} + \frac{1}{\lambda'_\alpha}$

(D) The wavelength of the $K_{\alpha}$ line is characteristic of the target material and is independent of the accelerating voltage $V$. Thus,$\lambda_{K_{\alpha}}$ remains constant.
The minimum wavelength of the continuous $X-ray$ spectrum is given by $\lambda_{\min} = \frac{hc}{eV}$.
When the voltage $V$ is halved to $V' = V/2$,the new minimum wavelength becomes $\lambda'_{\min} = \frac{hc}{e(V/2)} = 2\lambda_{\min}$.
The initial difference is $\Delta \lambda = \lambda_{K_{\alpha}} - \lambda_{\min}$.
The new difference is $\Delta \lambda' = \lambda_{K_{\alpha}} - 2\lambda_{\min}$.
Since $\lambda_{K_{\alpha}} > \lambda_{\min}$,we can write $\Delta \lambda' = \lambda_{K_{\alpha}} - 2\lambda_{\min} = (\lambda_{K_{\alpha}} - \lambda_{\min}) - \lambda_{\min} = \Delta \lambda - \lambda_{\min}$.
Since $\lambda_{\min} > 0$,it follows that $\Delta \lambda' < \Delta \lambda$.
Also,$2\Delta \lambda = 2\lambda_{K_{\alpha}} - 2\lambda_{\min}$. Comparing $\Delta \lambda' = \lambda_{K_{\alpha}} - 2\lambda_{\min}$ with $2\Delta \lambda$,we see that $\Delta \lambda' < 2\Delta \lambda$. Therefore,the difference becomes less than two times the original difference.

(A) The wavelength of the characteristic $X$-ray line,such as the ${K_\alpha }$ line $({\lambda _k})$,depends only on the material of the target and is independent of the accelerating voltage $(V)$.
The minimum wavelength of the continuous $X$-ray spectrum $({\lambda _c})$ is given by the Duane-Hunt law: ${\lambda _c} = \frac{hc}{eV}$.
From this relation,it is clear that ${\lambda _c}$ is inversely proportional to the accelerating voltage $(V)$.
As the accelerating voltage $(V)$ is increased,${\lambda _c}$ decreases,while ${\lambda _k}$ remains constant.
Therefore,the difference $({\lambda _K} - {\lambda _C})$ will increase.

(C) In $X-$ ray spectra,depending on the accelerating voltage and the target element,we may find sharp peaks superimposed on a continuous spectrum.
These sharp peaks occur at specific wavelengths that are unique to the target material used in the $X-$ ray tube.
These peaks arise due to electronic transitions within the atoms of the target material,where an electron from a higher energy shell drops into a vacancy in a lower energy shell,emitting a photon of specific energy.
Therefore,these peaks are known as characteristic radiations.

(C) According to Moseley's law,the frequency $\nu$ of a characteristic $X$-ray is related to the atomic number $Z$ of the target material by the equation:
$\sqrt{\nu} = a(Z - b)$
where $a$ and $b$ are constants.
Squaring both sides,we get:
$\nu = a^2(Z - b)^2$
This equation is of the form $\nu = k(Z - b)^2$,which represents a parabola opening upwards with its vertex at $(b, 0)$ on the $Z$-axis.
Comparing this with the given options,the graph that represents a parabolic relationship between $\nu$ and $Z$ is graph $C$.

(B) $1$. The minimum wavelength of the continuous $X$-ray spectrum is given by $\lambda_{\min} = \frac{hc}{eV}$.
$2$. From the graph,we observe that $\lambda_1 < \lambda_2$. Since $\lambda_{\min} \propto \frac{1}{V}$,a smaller minimum wavelength corresponds to a higher accelerating voltage. Therefore,$V_1 > V_2$.
$3$. The characteristic $K_{\alpha}$ line wavelength is determined by Moseley's law: $\sqrt{\nu} = a(Z - b)$,which implies $\nu \propto (Z - 1)^2$. Since $\nu = \frac{c}{\lambda}$,we have $\lambda \propto \frac{1}{(Z - 1)^2}$.
$4$. From the graph,the $K_{\alpha}$ peak for curve $1$ appears at a shorter wavelength than the $K_{\alpha}$ peak for curve $2$. Thus,$\lambda_{K\alpha, 1} < \lambda_{K\alpha, 2}$.
$5$. This implies $(Z_1 - 1)^2 > (Z_2 - 1)^2$,which means $Z_1 > Z_2$.
$6$. Combining these results,we get $V_1 > V_2$ and $Z_1 > Z_2$.

(A) In a Coolidge tube,the continuous $X-$ray spectrum is produced due to the deceleration of electrons hitting the target. The energy of the emitted photon $E = h\nu$ is derived from the kinetic energy of the incident electrons. The maximum frequency $\nu_{\max}$ corresponds to the case where the entire kinetic energy of an electron is converted into a single photon,given by $h\nu_{\max} = eV$,where $V$ is the accelerating potential. Since electrons can lose any fraction of their kinetic energy,the frequency $\nu$ of the emitted $X-$rays can range from $0$ up to a maximum value $\nu_{\max}$. The intensity $I$ is zero at $\nu = 0$ and $\nu = \nu_{\max}$,and it reaches a peak in between. This corresponds to the graph in option $A$.

(A) The short wavelength limit $\lambda _{\min }$ is given by the relation $\lambda _{\min } = \frac{hc}{eV}$.
Taking the logarithm on both sides,we get $\log \lambda _{\min } = \log \left( \frac{hc}{e} \right) - \log V$.
This can be rewritten as $\log \lambda _{\min } = - \log V + \log \left( \frac{hc}{e} \right)$.
This is an equation of a straight line of the form $y = mx + c$,where $y = \log \lambda _{\min }$,$x = \log V$,slope $m = -1$,and intercept $c = \log \left( \frac{hc}{e} \right)$.
Since the slope is negative,the graph is a straight line with a downward slope,which corresponds to curve $A$ in the figure.

(C) According to Moseley's law,the frequency $\nu$ of the characteristic $K_{\alpha}$ $X$-ray line is given by $\nu = cR(Z - b)^2$,where $b = 1$ for $K_{\alpha}$ lines.
Thus,$\nu \propto (Z - 1)^2$.
Since $\nu = \frac{c}{\lambda}$,we have $\frac{c}{\lambda} \propto (Z - 1)^2$.
This implies $\lambda \propto \frac{1}{(Z - 1)^2}$.
As $Z$ increases,$\lambda$ decreases rapidly. This relationship represents a rectangular hyperbola-like curve that decreases as $Z$ increases,which corresponds to curve $C$.

(B) $1$. In the characteristic $X$-ray spectrum,the $K_\alpha$ line has a longer wavelength and higher intensity compared to the $K_\beta$ line.
$2$. From the graph,the peak at $0.7 \ \mathring{A}$ is higher than the peak at $0.6 \ \mathring{A}$. Therefore,the $K_\alpha$ line corresponds to the peak at $0.7 \ \mathring{A}$.
$3$. If the energy of the incident electrons is increased,the minimum wavelength (cut-off wavelength) $\lambda_{min} = \frac{hc}{E}$ decreases,shifting the continuous spectrum towards the left. However,the characteristic peaks $(K_\alpha, K_\beta)$ depend only on the target material and remain at their fixed wavelengths. Thus,option $(c)$ is incorrect as the characteristic peaks do not shift,and option $(b)$ is the correct observation from the graph.

(B) For the ${K_{\alpha }}$ $X$-ray line,Moseley's law is given by:
$\frac{1}{\lambda} = R(Z - 1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = \frac{3R}{4}(Z - 1)^2$
Given $\lambda = 0.76 \, \mathring{A} = 0.76 \times 10^{-10} \, m$ and Rydberg constant $R \approx 1.097 \times 10^7 \, m^{-1}$.
Substituting the values:
$\frac{1}{0.76 \times 10^{-10}} = \frac{3}{4} \times 1.097 \times 10^7 \times (Z - 1)^2$
$1.315 \times 10^{10} = 0.82275 \times 10^7 \times (Z - 1)^2$
$(Z - 1)^2 = \frac{1.315 \times 10^{10}}{0.82275 \times 10^7} \approx 1598.3$
$Z - 1 \approx \sqrt{1600} = 40$
$Z = 41$.

(D) Fraunhofer lines are dark absorption lines observed in the solar spectrum caused by the absorption of specific wavelengths by the cooler gases in the Sun's atmosphere (the photosphere).
During a total solar eclipse,the Moon blocks the bright photosphere of the Sun.
This allows the light from the chromosphere,which is a hotter and thinner layer of the Sun's atmosphere,to be observed.
The chromosphere emits light through emission,which produces bright emission lines at the exact same wavelengths where the dark absorption lines (Fraunhofer lines) were previously observed.
Therefore,the dark Fraunhofer lines are replaced by bright,brilliant emission lines.

(D) The correct answer is $D$. Fraunhofer lines are produced by the absorption of light rays from the Sun by the gases and vapors present in the solar atmosphere. When white light from the photosphere passes through the cooler chromosphere,the atoms and molecules present in the chromosphere absorb specific wavelengths of light. This process results in the formation of dark lines in the otherwise continuous spectrum of the Sun,which are known as Fraunhofer lines. Therefore,they represent a line absorption spectrum.

(B) In an $X$-ray tube,when electrons are accelerated through a potential difference $V$,they possess a maximum kinetic energy $K_{max} = eV$.
When these electrons strike the target,they produce continuous $X$-rays due to Bremsstrahlung radiation.
The maximum energy of an emitted photon corresponds to the case where the electron loses all its kinetic energy in a single collision,given by $E_{max} = h\nu_{max} = \frac{hc}{\lambda_{min}} = eV$.
Thus,the minimum wavelength is $\lambda_{min} = \frac{hc}{eV}$.
Since the electron can lose any amount of energy from $0$ to $eV$ in various collisions,the emitted $X$-ray photons can have any wavelength from $\lambda_{min}$ to $\infty$.
Therefore,the range of wavelengths for continuous $X$-rays is $\lambda_{min}$ to $\infty$.

(A) Characteristic $X$-rays are produced due to electronic transitions within the target atom. When a high-energy incident electron strikes the target,it knocks out an electron from an inner shell (e.g.,$K$-shell). This creates a vacancy in that shell. To stabilize the atom,an electron from a higher energy shell (e.g.,$L$ or $M$ shell) drops down to fill this vacancy. The difference in energy between these two shells is released in the form of a photon,which corresponds to the characteristic $X$-ray emission.

(B) The intensity of $X$-rays depends on the number of electrons hitting the target,which is determined by the filament current,not the accelerating voltage. The accelerating voltage determines the minimum wavelength (maximum energy) of the $X$-rays. Therefore,if the voltage is doubled,the intensity of the $X$-rays remains constant.

(C) $X$-rays are high-energy electromagnetic waves that easily pass through soft tissues like the intestines without being significantly absorbed or scattered. Because they do not cast a clear shadow of the soft tissue structures on the photographic film,they cannot provide a useful diagnostic image of the intestines. To visualize the intestines,radiologists use contrast agents like barium sulfate,which absorbs $X$-rays and creates the necessary contrast.

(A) Characteristic $X$-rays are produced due to the transition of electrons from higher energy levels to lower energy levels within the atoms of the target material.
Since the energy levels of an atom are unique to the element,the energy (and thus the wavelength) of the characteristic $X$-rays emitted depends solely on the atomic number $(Z)$ of the target material.
Therefore,the characteristic $X$-rays depend on the nature of the target material.

(A) When high-speed electrons (cathode rays) strike a metal target with a high melting point and high atomic number,they are suddenly decelerated. This loss of kinetic energy is emitted in the form of high-frequency electromagnetic radiation known as $X$-rays.

(B) For the emission of $K$-series characteristic $X$-rays,the energy of the incident electrons $(E = eV)$ must be greater than the ionization energy $(E_i)$ required to knock out an electron from the $K$-shell.
Given the operating voltage $V = 15 \, kV$,the energy of the incident electrons is $E = 15 \, keV$.
Comparing this with the ionization energies:
For $Co$: $7.8 \, keV < 15 \, keV$ (Emission possible).
For $Cu$: $9.0 \, keV < 15 \, keV$ (Emission possible).
For $Mo$: $20.1 \, keV > 15 \, keV$ (Emission not possible).
Thus,only $Co$ and $Cu$ can emit $K$-series characteristic $X$-rays.
The minimum wavelength of continuous $X$-rays is given by $\lambda_{min} = \frac{hc}{eV}$. Since $V$ is the same for all,$\lambda_{min}$ is the same for all three metals.

(A) The structure of a solid,specifically the arrangement of atoms in a crystal lattice,is determined using $X$-ray diffraction.
Because the wavelength of $X$-rays is of the order of the interatomic spacing in solids (approximately $1 \ \mathring{A}$ or $10^{-10} \ \text{m}$),they undergo diffraction when interacting with the crystal lattice.
This phenomenon allows scientists to map the atomic positions within the solid,a technique known as $X$-ray crystallography.

(A) The shortest wavelength (cutoff wavelength) of $X$-rays produced in an $X$-ray tube is given by the Duane-Hunt law: $\lambda_{min} = \frac{hc}{eV}$.
Here,$h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $V$ is the accelerating potential difference (voltage) applied across the tube.
Since $h$,$c$,and $e$ are constants,the minimum wavelength $\lambda_{min}$ is inversely proportional to the applied voltage $V$.
Therefore,the shortest wavelength depends on the voltage applied across the tube.

(B) In an $X$-ray tube,the characteristic $X$-rays are produced due to the transitions of electrons between the inner shells of the atoms of the target material.
Since the energy levels of the inner shells depend on the atomic number $(Z)$ of the element,the wavelength of the characteristic $X$-rays depends solely on the nature of the target material.
Therefore,the correct option is $B$.

(C) $X$-rays are electromagnetic waves consisting of high-energy photons.
They do not carry any electric charge.
Since magnetic fields only exert a force on moving electric charges (Lorentz force) or magnetic dipoles,and $X$-rays possess neither,they remain unaffected by magnetic fields.
Therefore,$X$-rays will not deviate when passing through a strong magnetic field.

(B) $X$-rays are produced due to electronic transitions in the inner shells of an atom,making them an atomic property.
$\gamma$-rays are produced due to transitions within the nucleus during radioactive decay,making them a nuclear property.
Therefore,the fundamental difference is that $X$-rays originate from the electron cloud (atomic),while $\gamma$-rays originate from the nucleus.

(B) In an $X$-ray tube, the intensity of the emitted radiation depends on the filament current, while the minimum wavelength $(\lambda_{min})$ depends on the accelerating potential difference $(V)$.
$1$. The intensity of the $X$-rays remains constant if the filament current is not changed.
$2$. The minimum wavelength is given by the formula $\lambda_{min} = \frac{hc}{eV}$. As the potential difference $(V)$ increases, the minimum wavelength $(\lambda_{min})$ decreases.
Therefore, the correct observations are that the intensity remains constant and the minimum wavelength decreases, which corresponds to statements $(3)$ and $(4)$.

(A) In an $X$-ray tube,the intensity of the emitted $X$-ray beam is directly proportional to the number of electrons striking the target per unit time.
This number of electrons is determined by the thermionic emission from the filament.
Therefore,increasing the filament current increases the temperature of the filament,which in turn increases the thermionic emission of electrons.
Consequently,the intensity of the $X$-ray beam increases.

(B) In an $X$-ray spectrum,the $K$-series lines are produced when electrons transition from higher energy levels $(n=2, 3, 4, ...)$ to the $K$-shell $(n=1)$.
The energy of the emitted photon is given by $E = \frac{hc}{\lambda}$,which implies $\lambda = \frac{hc}{E}$.
The energy difference for the transitions are:
For $K_\alpha$: $E_\alpha = E_2 - E_1$
For $K_\beta$: $E_\beta = E_3 - E_1$
For $K_\gamma$: $E_\gamma = E_4 - E_1$
Since $E_4 > E_3 > E_2$,it follows that $E_\gamma > E_\beta > E_\alpha$.
Because wavelength is inversely proportional to energy,we have $\lambda_\alpha > \lambda_\beta > \lambda_\gamma$.

(A) $X$-rays are electromagnetic waves with wavelengths in the range of $0.01 \ nm$ to $10 \ nm$.
Human eyes are only sensitive to the visible spectrum,which ranges approximately from $380 \ nm$ to $750 \ nm$.
Since the wavelength of $X$-rays is much shorter than the visible spectrum,they cannot be detected by the human eye.
Therefore,$X$-rays are invisible to the human eye.

(C) To produce characteristic $X$-rays,an incident electron must have enough kinetic energy to knock out an electron from the innermost shell ($K$-shell) of the target atom.
The energy required to remove this electron is equal to its binding energy,which is $40 \, keV$.
Therefore,the kinetic energy $K$ of the incident electron must be greater than the binding energy $E_b$:
$K > E_b$
Since the kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K = eV$,we have:
$eV > 40 \, keV$
Dividing both sides by $e$,we get:
$V > 40 \, kV$
Thus,the potential difference must be greater than $40 \, kV$ to ionize the atom and produce characteristic $X$-rays.

(C) An $X$-ray tube requires a very high vacuum to prevent the ionization of residual gas molecules by accelerated electrons,which would cause electrical discharge and damage the tube. The operating pressure inside an $X$-ray tube is typically maintained at a very low level,approximately $10^{-5} \, mm$ to $10^{-6} \, mm$ of $Hg$. Among the given options,$10^{-6} \, mm$ of $Hg$ is the correct order of magnitude.

(C) Characteristic $X$-rays are produced by transitions between inner electron shells (like $K$ or $L$ shells) in atoms with high atomic numbers $(Z)$.
In a hydrogen atom,there is only one electron and the energy difference between its energy levels is very small (in the order of $eV$).
$X$-rays require energy transitions in the range of $keV$,which is not possible in a hydrogen atom due to its low atomic number and small energy gap between levels.

(B) Continuous $X$-rays are produced due to the deceleration (or retardation) of incident high-speed electrons as they interact with the electric field of the nuclei in the target material. This process is known as Bremsstrahlung radiation. As the electrons lose kinetic energy,the energy is emitted in the form of continuous $X$-ray photons.