X-Rays Questions in English

(D) According to Moseley's law for $K_{\alpha}$ $X$-rays: $\frac{1}{\lambda} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3R}{4}(Z-1)^2$.
For metal $A$: $\frac{1875R}{4} = \frac{3R}{4}(Z_A-1)^2 \implies (Z_A-1)^2 = 625 \implies Z_A-1 = 25 \implies Z_A = 26$.
For metal $B$: $675R = \frac{3R}{4}(Z_B-1)^2 \implies (Z_B-1)^2 = 900 \implies Z_B-1 = 30 \implies Z_B = 31$.
The number of elements between $A$ and $B$ is given by $Z_B - Z_A - 1 = 31 - 26 - 1 = 4$.

(B) The de-Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.
The shortest wavelength of the $X$-rays produced (cut-off wavelength) is given by $\lambda_2 = \frac{hc}{eV}$.
The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2meV}} \times \frac{eV}{hc} = \frac{1}{c} \sqrt{\frac{eV}{2m}} = \frac{1}{c} \sqrt{\frac{V}{2} \cdot \frac{e}{m}}$.
Substituting the given values: $V = 10^4\, V$,$\frac{e}{m} = 1.8 \times 10^{11}\, C\, kg^{-1}$,and $c = 3 \times 10^8\, m/s$.
$\frac{\lambda_1}{\lambda_2} = \frac{1}{3 \times 10^8} \sqrt{\frac{10^4}{2} \times 1.8 \times 10^{11}} = \frac{1}{3 \times 10^8} \sqrt{0.9 \times 10^{15}} = \frac{1}{3 \times 10^8} \sqrt{9 \times 10^{14}} = \frac{3 \times 10^7}{3 \times 10^8} = 0.1$.

(D) $X$-rays have a wavelength on the order of the inter-atomic spacing in solid crystals $(10^{-10} \ m)$.
Because the wavelength of $X$-rays is comparable to the inter-atomic distances,they undergo diffraction when interacting with the crystal lattice.
Therefore,$X$-rays are the most suitable electromagnetic waves for investigating the structure of solids.

(C) The minimum wavelength $(\lambda_{\min})$ produced in an $X$-ray tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400}{V(\text{in volts})} \mathring{A}$ (approximately).
Using the given potential difference $V = 50\, kV = 50,000\, V$:
$\lambda_{\min} = \frac{12375}{50,000} \mathring{A}$.
$\lambda_{\min} = 0.2475 \mathring{A}$.
Rounding to the nearest given option,we get $\lambda_{\min} \approx 0.25 \mathring{A}$.

(D) Characteristic $X$-rays are produced when an electron from an outer shell fills a vacancy in an inner shell (such as the $K$-shell) of an atom.
When a high-energy electron strikes an atom,it can eject an inner-shell electron,creating a vacancy.
An electron from a higher energy level then transitions to fill this vacancy,emitting a photon with energy equal to the difference between the two energy levels.
This process is fundamentally related to the electronic transitions within the atom,which is also the mechanism involved in $K$-electron capture,where a nucleus captures an inner-shell electron,leading to a vacancy that is subsequently filled by an outer-shell electron,resulting in characteristic $X$-ray emission.

(A) The energy of an electron accelerated through a potential difference $V$ is $E = eV$.
For the minimum wavelength (maximum energy photon),all kinetic energy of the electron is converted into a single photon: $E = \frac{hc}{\lambda_{\min}}$.
Equating the two: $eV = \frac{hc}{\lambda_{\min}}$,which gives $\lambda_{\min} = \frac{hc}{eV}$.
Given $V = 20 \, kV = 20 \times 10^3 \, V$.
Using the constant $hc \approx 12400 \, eV \cdot \mathring{A}$,we get:
$\lambda_{\min} = \frac{12400 \, eV \cdot \mathring{A}}{20 \times 10^3 \, eV} = \frac{12400}{20000} \, \mathring{A} = 0.62 \, \mathring{A}$.

(D) The most energetic $X$-rays correspond to the minimum wavelength $(\lambda_{\min})$ produced when the entire kinetic energy of the electron is converted into a single photon.
Using the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Given: $V = 40 \ keV = 40 \times 10^3 \ V$,$h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $e = 1.6 \times 10^{-19} \ C$.
Alternatively,using the shortcut $\lambda_{\min} \approx \frac{12400 \ \mathring{A} \cdot eV}{E \text{ (in } eV)} = \frac{12400}{40000} \ \mathring{A}$.
$\lambda_{\min} = 0.31 \ \mathring{A}$.

(A) All electromagnetic waves travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$ in a vacuum.
Since $X-$rays are a part of the electromagnetic spectrum,they are electromagnetic waves.
Therefore,the Assertion is correct because the Reason correctly identifies their nature as electromagnetic waves,which dictates their speed.

(B) The energy of an $X$-ray photon is given by $E = \frac{hc}{\lambda}$.
For the minimum wavelength $\lambda_{min}$,the accelerating voltage $V$ is given by $eV = \frac{hc}{\lambda_{min}}$.
Substituting the values $h = 6.63 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^8 \, m/s$,$e = 1.6 \times 10^{-19} \, C$,and $\lambda = 10^{-11} \, m$:
$V = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 10^{-11}}$
$V = \frac{19.89 \times 10^{-26}}{1.6 \times 10^{-30}} = 12.43 \times 10^4 \, V = 124.3 \, kV$.
Since the wavelength must be at least $10^{-11} \, m$,the energy of the electrons must be sufficient to produce this wavelength. Therefore,the accelerating voltage must be at least $124.3 \, kV$ to produce wavelengths equal to or shorter than this limit. However,in the context of the question asking for the condition to produce this specific minimum wavelength,the voltage must be at least $124.2 \, kV$ (using $hc \approx 1242 \, eV \cdot nm$). Thus,$V \geq 124.2 \, kV$.

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
For the first photon:
$E_1 = 2.5\,\text{eV}$ and $\lambda_1 = 5000\,\mathring{A}$.
So,$2.5 = \frac{hc}{5000\,\mathring{A}} \implies hc = 2.5 \times 5000\,\text{eV} \cdot \mathring{A}$.
For the $X$-ray photon:
$E_2 = \frac{hc}{\lambda_2}$ where $\lambda_2 = 1\,\mathring{A}$.
Substituting the value of $hc$:
$E_2 = \frac{2.5 \times 5000}{1} = 12500\,\text{eV}$.
Thus,the energy is $2.5 \times 5000\,\text{eV}$.

(D) The Assertion is incorrect because while soft and hard $X-$ rays differ in frequency,they both travel with the same velocity,which is the speed of light $(c \approx 3 \times 10^8 \ m/s)$ in a vacuum.
The Reason is correct because hard $X-$ rays have higher energy and frequency,which gives them greater penetrating power compared to soft $X-$ rays.
Therefore,the Assertion is incorrect and the Reason is correct.

(B) According to Moseley's law,the energy of a photon emitted during an electronic transition in an atom is given by $\Delta E = Rch(Z-1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the $K_{\alpha}$ line,the transition is from $n_2 = 2$ to $n_1 = 1$. Thus,$\Delta E_{\alpha} \propto \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \left( 1 - \frac{1}{4} \right) = \frac{3}{4} = 0.75$.
For the $K_{\beta}$ line,the transition is from $n_2 = 3$ to $n_1 = 1$. Thus,$\Delta E_{\beta} \propto \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = \left( 1 - \frac{1}{9} \right) = \frac{8}{9} \approx 0.889$.
Since $\Delta E_{\alpha} < \Delta E_{\beta}$,the ratio $\frac{\Delta E_{\alpha}}{\Delta E_{\beta}}$ is less than $1$.

(N/A) The maximum frequency of $X$-rays is produced when the entire kinetic energy of the electron is converted into a single photon. The energy of the photon is given by $E = h \nu_{\max} = eV$.
Thus,$\nu_{\max} = \frac{eV}{h}$.
Substituting the values: $\nu_{\max} = \frac{(1.6 \times 10^{-19} \; C) \times (30 \times 10^3 \; V)}{6.63 \times 10^{-34} \; J \cdot s} = 7.24 \times 10^{18} \; Hz$.
$(b)$ The minimum wavelength is related to the maximum frequency by $\lambda_{\min} = \frac{c}{\nu_{\max}}$.
Substituting the values: $\lambda_{\min} = \frac{3 \times 10^8 \; m/s}{7.24 \times 10^{18} \; Hz} \approx 0.0414 \times 10^{-9} \; m = 0.0414 \; nm$.

(N/A) The wavelength produced by an $X$-ray tube is $\lambda = 0.45\,\mathring{A} = 0.45 \times 10^{-10}\,\text{m}$.
Planck's constant,$h = 6.626 \times 10^{-34}\,\text{J s}$.
Speed of light,$c = 3 \times 10^{8}\,\text{m/s}$.
The maximum energy of a photon is given by $E = \frac{hc}{\lambda}$.
Converting to electron-volts $(eV)$: $E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.45 \times 10^{-10} \times 1.602 \times 10^{-19}}\,\text{eV} \approx 27.6 \times 10^{3}\,\text{eV} = 27.6\,\text{keV}$.
Therefore,the maximum energy of an $X$-ray photon is $27.6\,\text{keV}$.
$(b)$ The accelerating voltage provides kinetic energy to the electrons to produce $X$-rays. To obtain an $X$-ray photon of $27.6\,\text{keV}$,the incident electrons must possess at least $27.6\,\text{keV}$ of kinetic energy.
Hence,an accelerating voltage of the order of $30\,\text{kV}$ is required for producing these $X$-rays.

(A) $X-$rays were discovered by the German physicist Wilhelm Conrad Roentgen in $1895$. He observed that these rays were produced when high-energy electrons struck a metal target in a vacuum tube. For this groundbreaking discovery,he was awarded the first Nobel Prize in Physics in $1901$.

(A) $X$-rays are produced when high-energy electrons are bombarded on a metal target. Their wavelength ranges from $10^{-8} \text{ m}$ to $10^{-13} \text{ m}$.
Uses of $X$-rays:
$(i)$ Used in medical science to detect bone fractures and treat certain types of cancer.
$(ii)$ Used in industrial applications to detect cracks and flaws in metal components.
Gamma rays are emitted during nuclear processes and from certain radioactive nuclei. They have the highest frequency in the electromagnetic spectrum,with a wavelength range of $10^{-10} \text{ m}$ to $10^{-14} \text{ m}$.
Uses of gamma rays:
$(i)$ Used in medical science to destroy cancer cells (radiotherapy).
$(ii)$ Used in the food industry for sterilization and to kill bacteria.

(A) $X-$rays are produced when high-energy electrons,accelerated by a high potential difference,strike a metal target of high atomic number (such as tungsten).
When these fast-moving electrons interact with the atoms of the target,they undergo rapid deceleration (braking radiation or Bremsstrahlung) or knock out inner-shell electrons,causing transitions that emit high-frequency electromagnetic radiation known as $X-$rays.

(A) The shortest wavelength (cut-off wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
Given that the accelerating potential $V = 1.24 \times 10^{6} \, V$,we can use the simplified relation: $\lambda_{\min} \approx \frac{1240}{V \text{ (in kV)}} \, nm$.
Here,$V = 1.24 \times 10^{6} \, V = 1240 \times 10^{3} \, kV$.
Substituting the values: $\lambda_{\min} = \frac{1240}{1.24 \times 10^{6}} \, nm = 10^{-3} \, nm$.

(C) The energy of an $X$-ray photon is given by $E = \frac{hc}{\lambda}$.
According to the problem,the mass $m$ of a fictitious particle having the same energy is given by $E = mc^2$.
Equating the two expressions: $\frac{hc}{\lambda} = mc^2$.
Solving for $m$: $m = \frac{h}{c\lambda}$.
Given $\lambda = 10 \, \mathring{A} = 10 \times 10^{-10} \, \text{m} = 10^{-9} \, \text{m}$.
Substituting the value of $c = 3 \times 10^8 \, \text{m/s}$:
$m = \frac{h}{(3 \times 10^8) \times 10^{-9}} = \frac{h}{3 \times 10^{-1}} = \frac{h}{0.3} = \frac{10h}{3}$.
Comparing this with the given expression $\frac{x}{3} h$,we get $x = 10$.

(D) The de-Broglie wavelength of the electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.
The kinetic energy $K$ of the electron is $K = \frac{p^2}{2m}$.
Substituting $p = \frac{h}{\lambda}$,we get $K = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$.
In an $X$-ray tube,the maximum energy of the emitted $X$-ray photon (corresponding to the cut-off wavelength $\lambda_c$) is equal to the kinetic energy of the incident electron: $K = \frac{hc}{\lambda_c}$.
Equating the two expressions for $K$: $\frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_c}$.
Solving for $\lambda_c$: $\lambda_c = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

(D) The energy of the $K_{\alpha}$ $X$-ray photon is given by the difference in energy between the $K$ shell and the $L$ shell: $E_{K_{\alpha}} = E_{K} - E_{L}$.
The energy of the photon is calculated as $E_{K_{\alpha}} = \frac{hc}{\lambda}$.
Given $h = 4.14 \times 10^{-15} \, eVs$ and $c = 3 \times 10^{8} \, ms^{-1}$,the product $hc = 12.42 \times 10^{-7} \, eV \cdot m$.
Substituting the wavelength $\lambda = 0.071 \times 10^{-9} \, m$:
$E_{K_{\alpha}} = \frac{12.42 \times 10^{-7}}{0.071 \times 10^{-9}} \, eV \approx 17493 \, eV \approx 17.5 \, keV$.
We are given the energy of the atom with a $K$ electron removed as $E_{K} = 27.5 \, keV$.
Using the relation $E_{L} = E_{K} - E_{K_{\alpha}}$:
$E_{L} = 27.5 \, keV - 17.5 \, keV = 10 \, keV$.

(D) The continuous $X$-ray spectrum is produced when a high-speed electron (often called a projectile or incident electron) is decelerated by the electric field of a target atom's nucleus. This process is known as Bremsstrahlung or braking radiation.
During this interaction,the electron loses kinetic energy,which is emitted in the form of an $X$-ray photon. The energy of the emitted photon is given by $\Delta K = hf$,where $\Delta K$ is the change in kinetic energy of the electron.
This process is essentially the inverse of the photoelectric effect. In the photoelectric effect,a photon is absorbed to eject an electron from a material. In the production of continuous $X$-rays,an electron is decelerated to emit a photon. Therefore,it is referred to as the inverse photoelectric effect.

(C) $X$-rays are ionizing radiations. When the electroscope is exposed to $X$-rays,they ionize the air molecules inside the electroscope enclosure.
These ions (positive and negative) are attracted to the charged gold leaves.
The charges of opposite polarity to that on the leaves neutralize the charge on the leaves.
As the net charge on the leaves decreases,the electrostatic repulsion between them decreases,causing the leaves to collapse.

(A) The Compton shift $\Delta \lambda$ is given by the formula: $\Delta \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$.
Here, the Compton wavelength $\lambda_c = \frac{h}{m_0 c} \approx 2.43 \times 10^{-12} \, m = 2.43 \, pm$.
The scattering angle $\phi = 85^{\circ}$.
Substituting the values: $\Delta \lambda = 2.43 \times (1 - \cos 85^{\circ})$.
Since $\cos 85^{\circ} \approx 0.087$, we have $\Delta \lambda = 2.43 \times (1 - 0.087) = 2.43 \times 0.913 \approx 2.218 \, pm$.
Rounding to the nearest given option, the Compton shift is approximately $2.2 \, pm$.

(C) The minimum wavelength (cut-off wavelength) of $X$-rays is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400}{V} \text{ Å}$.
From this relation, we can see that $\lambda_{\min} \propto \frac{1}{V}$.
Given that the initial potential is $V$ and the initial wavelength is $\lambda$, we have $\lambda = \frac{k}{V}$ (where $k$ is a constant).
When the potential is changed to $V' = 2V$, the new minimum wavelength $\lambda'$ will be:
$\lambda' = \frac{k}{V'} = \frac{k}{2V} = \frac{1}{2} \left( \frac{k}{V} \right) = \frac{\lambda}{2}$.

(C) The minimum wavelength $(\lambda_{\min})$ of $X$-rays produced by an $X$-ray tube is given by the Duane-Hunt law:
$\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \text{ } \mathring{A} \cdot V}{V}$
Given $\lambda_{\min} = 0.45 \mathring{A}$.
Substituting the value into the formula:
$0.45 = \frac{12400}{V}$
$V = \frac{12400}{0.45}$
$V \approx 27555 \text{ } V$
Comparing this with the given options, the closest value is $27500 \text{ } V$.

(D) $X$-rays are electromagnetic waves that carry momentum and energy.
$1$. When $X$-rays strike a material,they exert radiation pressure,which results in a force on the material.
$2$. They transfer energy to the material through absorption or scattering processes.
$3$. Due to the photoelectric effect,$X$-rays can cause the emission of electrons from the surface of the material.
Therefore,all the given statements are correct.

(A) The penetrating power of $X$-rays is directly proportional to their energy.
According to the relation $E = eV$,where $E$ is the energy of the $X$-ray photon,$e$ is the elementary charge,and $V$ is the accelerating potential applied across the $X$-ray tube.
As the accelerating potential $V$ increases,the energy of the $X$-rays increases,which in turn increases their penetrating power.
Therefore,the correct option is $A$.

(B) The cut-off wavelength (also known as the minimum wavelength) of $X$-rays produced in a Coolidge tube is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400}{V} \text{ Å}$.
From this relation, it is clear that $\lambda_{\min} \propto \frac{1}{V}$.
If the potential difference $V$ is doubled (i.e., $V' = 2V$), the new cut-off wavelength $\lambda'_{\min}$ becomes:
$\lambda'_{\min} = \frac{12400}{2V} = \frac{1}{2} \left( \frac{12400}{V} \right) = \frac{1}{2} \lambda_{\min}$.
Therefore, the cut-off wavelength is halved.

(C) According to Moseley's Law,the frequency $v$ of characteristic $X$-rays is given by $\sqrt{v} = a(Z - b)$,where $Z$ is the atomic number and $b$ is the screening constant. For $K_\alpha$ $X$-rays,$b = 1$.
For the first element with $Z_1 = 31$,the frequency is $v_1 = v$:
$\sqrt{v} = a(31 - 1) = 30a \quad \dots (i)$
For the second element with $Z_2 = 51$,let the frequency be $v_2$:
$\sqrt{v_2} = a(51 - 1) = 50a \quad \dots (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{\sqrt{v_2}}{\sqrt{v}} = \frac{50a}{30a} = \frac{5}{3}$
Squaring both sides:
$\frac{v_2}{v} = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$
Therefore,$v_2 = \frac{25}{9} v$.

(C) The $X$-ray spectrum produced by an $X$-ray tube consists of a continuous spectrum of wavelengths.
When high-speed electrons strike the target,they lose energy in the form of $X$-ray photons.
The maximum energy of an $X$-ray photon corresponds to the kinetic energy of the incident electron,given by $E_{max} = eV = \frac{hc}{\lambda_{min}}$.
This implies that there is a minimum wavelength $\lambda_{min} = \frac{hc}{eV}$ below which no $X$-rays are produced.
Since electrons can lose any fraction of their kinetic energy,the resulting $X$-ray beam contains a continuous range of wavelengths starting from $\lambda_{min}$ up to infinity (theoretically).
Therefore,the beam has all wavelengths greater than a certain minimum wavelength.

(B) According to Moseley's Law for $K_\alpha$ $X$-rays,the frequency $f$ is given by $\sqrt{f} = a(Z - b)$,where $a = \sqrt{\frac{3}{4} R c}$ and $b = 1$.
Substituting the values,$\sqrt{f} = \sqrt{\frac{3}{4} \times 1.097 \times 10^7 \times 3 \times 10^8} \times (41 - 1)$.
$\sqrt{f} \approx 4.97 \times 10^7 \times 40 = 1.988 \times 10^9 \text{ Hz}^{1/2}$.
Squaring both sides,$f \approx 3.95 \times 10^{18} \text{ Hz}$.
Using $\lambda = \frac{c}{f}$,where $c = 3 \times 10^8 \text{ m/s}$.
$\lambda = \frac{3 \times 10^8}{3.95 \times 10^{18}} \approx 0.76 \times 10^{-10} \text{ m} = 0.76 \,\mathring A$.

(B) The velocity $v$ of the photoelectrons is given by the magnetic force providing the centripetal force: $evB = m\frac{v^2}{R} \implies v = \frac{eBR}{m}$.
The kinetic energy $K$ of the photoelectrons is $K = \frac{1}{2}mv^2 = \frac{e^2B^2R^2}{2m}$.
Substituting the values: $e = 1.6 \times 10^{-19}\; C$,$B = 2 \times 10^{-2}\; T$,$R = 23 \times 10^{-3}\; m$,$m = 9.1 \times 10^{-31}\; kg$:
$K = \frac{(1.6 \times 10^{-19})^2 \times (2 \times 10^{-2})^2 \times (23 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} \approx 2.97 \times 10^{-15}\; J$.
Converting to $keV$: $K = \frac{2.97 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 18.56\; keV$.
The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{12.4\; keV \cdot \mathring{A}}{0.50\; \mathring{A}} = 24.8\; keV$.
The binding energy $BE$ is given by $BE = E - K = 24.8\; keV - 18.6\; keV = 6.2\; keV$.

(B) When a metal target is bombarded with high-energy electrons,the electrons undergo rapid deceleration upon interacting with the target atoms. This loss of kinetic energy is emitted in the form of electromagnetic radiation known as $X$-rays. Therefore,the correct answer is $X$-rays.

(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K = eV$,where $e$ is the charge of the electron.
When this electron strikes the target,the maximum energy of the emitted $X$-ray photon is equal to the kinetic energy of the electron,i.e.,$E_{\max} = h\nu_{\max} = eV$.
Since the frequency $\nu = \frac{c}{\lambda}$,the minimum wavelength $\lambda_{\min}$ corresponds to the maximum energy:
$\lambda_{\min} = \frac{hc}{E_{\max}} = \frac{hc}{eV}$.
Since $h$,$c$,and $e$ are constants,we have $\lambda_{\min} \propto \frac{1}{V}$.
Therefore,the correct option is $C$.

(B) The characteristic $X$-rays are produced due to electronic transitions between the inner shells of the target atoms. The energy of these transitions depends only on the atomic number $(Z)$ of the target material, not on the accelerating potential of the electrons. Thus, $Statement-1$ is True.
When an electron beam strikes the target, it loses kinetic energy, which is emitted as $X$-ray photons (Bremsstrahlung and characteristic). This is a general physical process occurring in the tube. Thus, $Statement-2$ is True.
However, $Statement-2$ describes the general production of $X$-rays, whereas $Statement-1$ refers specifically to the independence of characteristic $X$-ray wavelengths from the accelerating potential. Therefore, $Statement-2$ is not the specific explanation for $Statement-1$.

(B) The cut-off wavelength $(\lambda_{\text{min}})$ of continuous $X$-rays is given by the formula $\lambda_{\text{min}} = \frac{hc}{eV}$, where $V$ is the accelerating potential difference.
This formula shows that the cut-off wavelength depends only on the accelerating voltage $(V)$ and is independent of the atomic number $(Z)$ of the target material.
Therefore, the statement that the cut-off wavelength depends on the atomic number of the target is $WRONG$.

$(A)$ $1$. The cut-off wavelength $(\lambda_{\min})$ is given by the formula: $\lambda_{\min} = \frac{hc}{eV}$.
$2$. Since $V$ is increased to $2V$, the new cut-off wavelength $\lambda'_{\min} = \frac{hc}{e(2V)} = \frac{1}{2} \lambda_{\min}$. Thus, the cut-off wavelength reduces to half.
$3$. The wavelengths of characteristic $X$-rays depend only on the target material, not on the accelerating potential $V$ or the filament current $I$. Therefore, they remain unchanged.
$4$. The intensity of $X$-rays is directly proportional to the number of electrons hitting the target per unit time, which is determined by the filament current $I$. Since $I$ is decreased to $I/2$, the number of electrons hitting the target decreases, leading to a decrease in the intensity of all $X$-rays.
$5$. Combining these, the cut-off wavelength reduces to half, and the intensities decrease. Thus, statements $(A)$ and $(C)$ are correct.

(B) According to Moseley's law,the frequency $\nu$ of the $K_{\alpha}$ $X$-ray line is given by $\sqrt{\nu} = a(Z - b)$,where $b = 1$ for the $K_{\alpha}$ transition.
Since $\nu = \frac{c}{\lambda}$,we have $\sqrt{\frac{c}{\lambda}} = a(Z - 1)$,which implies $\frac{1}{\sqrt{\lambda}} \propto (Z - 1)$.
Therefore,the ratio of the wavelengths is given by $\frac{\sqrt{\lambda_{Mo}}}{\sqrt{\lambda_{Cu}}} = \frac{Z_{Cu} - 1}{Z_{Mo} - 1}$.
Squaring both sides,we get $\frac{\lambda_{Mo}}{\lambda_{Cu}} = \left( \frac{29 - 1}{42 - 1} \right)^2 = \left( \frac{28}{41} \right)^2$.
Thus,the ratio $\frac{\lambda_{Cu}}{\lambda_{Mo}} = \left( \frac{41}{28} \right)^2 = \frac{1681}{784} \approx 2.144$.
Therefore,the ratio is close to $2.14$.

(A) The wavelength of the $K_\alpha$ line is given by Moseley's law: $\frac{1}{\lambda_{K_\alpha}} = R(Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R(Z-1)^2 \left( \frac{3}{4} \right)$.
The cut-off wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{eV}$,where $V$ is the accelerating potential of the electron beam.
Given the ratio $r = \frac{\lambda_{K_\alpha}}{\lambda_0} = 2$ for $Z_1 = 46$.
From the expression for $\lambda_{K_\alpha}$,we see that $\lambda_{K_\alpha} \propto \frac{1}{(Z-1)^2}$.
Since the accelerating potential $V$ is the same,$\lambda_0$ remains constant. Therefore,$r \propto \frac{1}{(Z-1)^2}$.
Thus,$\frac{r_2}{r_1} = \frac{(Z_1-1)^2}{(Z_2-1)^2}$.
Substituting the values: $\frac{r_2}{2} = \frac{(46-1)^2}{(41-1)^2} = \frac{45^2}{40^2} = \left( \frac{9}{8} \right)^2 = \frac{81}{64} = 1.2656$.
$r_2 = 2 \times 1.2656 = 2.5312 \approx 2.53$.

(A) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.
Since the kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the de-Broglie equation: $\lambda = \frac{h}{\sqrt{2mE}}$.
Squaring both sides: $\lambda^2 = \frac{h^2}{2mE}$,which gives $E = \frac{h^2}{2m\lambda^2}$.
The cut-off wavelength $\lambda_0$ of the emitted $X$-rays corresponds to the maximum energy of the photon,which is equal to the kinetic energy $E$ of the incident electron: $E = \frac{hc}{\lambda_0}$.
Equating the two expressions for $E$: $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.
Solving for $\lambda_0$: $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

(D) The electromagnetic spectrum classifies waves based on their wavelength and frequency. $X$-rays are high-energy electromagnetic waves that occupy the region between ultraviolet rays and gamma rays. The typical wavelength range for $X$-rays is approximately $1 \, nm$ to $10^{-3} \, nm$ ($0.1 \, \text{\AA}$ to $10 \, \text{\AA}$). Therefore, option $D$ is the correct answer.

(D) $X$-rays are high-energy electromagnetic radiations produced by transitions between inner electron shells (like $K$-shell to $L$-shell) 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 $eV$).
$X$-ray emission requires energy transitions in the order of $keV$.
Therefore,a hydrogen atom cannot emit $X$-rays because the energy levels are too close to each other to produce such high-energy photons.

(C) When a beam of high-energy electrons strikes a metal target,the sudden deceleration of electrons results in the emission of high-frequency electromagnetic radiation known as $X$-rays.

(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: $\lambda_{min} = \frac{hc}{eV}$.
Substituting the values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,$e = 1.6 \times 10^{-19} \ C$,and $V = 20 \times 10^3 \ V$.
$\lambda_{min} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{(1.6 \times 10^{-19}) \times (20 \times 10^3)} \ m$.
$\lambda_{min} = \frac{19.89 \times 10^{-26}}{3.2 \times 10^{-15}} \ m = 6.215 \times 10^{-11} \ m$.
Since $1 \ Å = 10^{-10} \ m$,we have $\lambda_{min} = 0.6215 \times 10^{-10} \ m = 0.62 \ Å$.

(C) The structure of solids is investigated using $X$-ray diffraction. Since the wavelength of $X$-rays is of the order of the interatomic spacing in crystals (approximately $1 \ \text{\AA}$), they are diffracted by the atomic planes in a solid. This phenomenon is used to determine the crystal structure of materials.

(B) The shortest wavelength (also known as the cut-off wavelength,$\lambda_{min}$) of $X-$rays emitted from 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,$\lambda_{min}$ depends inversely on the voltage $V$ applied to the tube.
Therefore,the shortest wavelength depends only on the voltage applied to the tube.

(B) Wilhelm Roentgen,a Professor of Physics in Wurzburg,Bavaria,discovered $X$-rays in $1895$ accidentally while testing whether cathode rays could pass through glass.
$X$-rays are electromagnetic radiation of extremely short wavelength and high frequency,with wavelengths ranging from about $10^{-8} \ m$ to $10^{-12} \ m$ and corresponding frequencies from about $10^{16} \ Hz$ to $10^{20} \ Hz$.

(A) Given, $Z=31$ and $a=5 \times 10^7 \, \text{Hz}^{1/2}$.
According to Moseley's law for $K_{\alpha}$ lines, $\sqrt{\nu} = a(Z-1)$.
Squaring both sides, we get $\nu = a^2(Z-1)^2$.
Substituting the values: $\nu = (5 \times 10^7)^2 \times (31-1)^2$.
$\nu = 25 \times 10^{14} \times 30^2 = 25 \times 10^{14} \times 900 = 2.25 \times 10^{18} \, \text{Hz}$.
We know that $\lambda = \frac{c}{\nu}$, where $c = 3 \times 10^8 \, \text{m/s}$.
$\lambda = \frac{3 \times 10^8}{2.25 \times 10^{18}} = 1.33 \times 10^{-10} \, \text{m}$.
Since $1 \, \mathring{A} = 10^{-10} \, \text{m}$, the wavelength is $1.33 \, \mathring{A}$.