X-Rays Questions in English

(B) When high-speed cathode rays (electrons) strike a target material (anode) with a high atomic weight (high $Z$),the electrons are rapidly decelerated by the strong electric field of the target nuclei.
This sudden deceleration causes the emission of electromagnetic radiation in the form of $X-rays$.
This is the fundamental principle behind the production of $X-rays$ in a Coolidge tube.

(C) The energy $E$ of a photon is given by the relation $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
Given,for $\lambda_1 = 5000 \, \mathring{A}$,$E_1 = 2.5 \, eV$.
We need to find the energy $E_2$ for $\lambda_2 = 1 \, \mathring{A}$.
Using the proportionality $E_1 \lambda_1 = E_2 \lambda_2$,we get:
$E_2 = E_1 \times \frac{\lambda_1}{\lambda_2}$
$E_2 = 2.5 \times \frac{5000}{1} \, eV$
$E_2 = 2.5 \times 5000 \, eV$.

(C) $X-$rays are electromagnetic waves with a wavelength range typically between $0.1 \, \mathring{A}$ and $100 \, \mathring{A}$.
Among the given options,$1 \, \mathring{A}$ falls within this range.
Therefore,the correct option is $C$.

(A) The penetrating power of $X$-rays is inversely proportional to their wavelength.
Shorter wavelengths correspond to higher energy and higher frequency $X$-rays,which possess greater penetrating power.
Comparing the given options,the wavelength of $2 \ \mathring{A}$ is the shortest.
Therefore,$X$-rays with a wavelength of $2 \ \mathring{A}$ will penetrate the metal block the most.

(D) The minimum wavelength (cutoff wavelength) 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 of constants $h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $e = 1.602 \times 10^{-19} \ C$,the formula simplifies to:
$\lambda_{\min} (\text{in } \mathring{A}) \approx \frac{12400}{V (\text{in } V)}$ or $\lambda_{\min} (\text{in } \mathring{A}) \approx \frac{12.4}{V (\text{in } kV)}$
Given $\lambda_{\min} = 0.3094 \ \mathring{A}$,we have:
$0.3094 = \frac{12400}{V}$
$V = \frac{12400}{0.3094} \approx 40077 \ V = 40.077 \ kV$
Rounding to the nearest given option,we get $V \approx 40 \ kV$.

(B) Radiotherapy is a medical treatment that uses high-energy radiation,such as $X$-rays,to kill cancer cells or shrink tumors.
By using controlled exposure,the radiation damages the $DNA$ of the cancer cells,preventing them from dividing and growing.
Therefore,the correct application of $X$-rays in this context is to treat cancer.

(A) The energy difference between the energy levels of a hydrogen atom is given by $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
For the hydrogen atom,the maximum energy transition (from $n = \infty$ to $n = 1$) is only $13.6 \text{ eV}$.
$X$-rays are high-energy electromagnetic radiations typically ranging from $100 \text{ eV}$ to $100 \text{ keV}$.
Since the energy released during electronic transitions in a hydrogen atom is significantly lower than the energy required to produce $X$-rays,it cannot emit $X$-rays.
Therefore,the correct option is $A$.

(B) The correct answer is $B$. $X$-rays were discovered by Wilhelm Conrad Roentgen,a German physicist,on November $8, 1895$. While working with a cathode-ray tube in his laboratory,Roentgen observed a fluorescent glow of crystals on a table near his tube,which led to the discovery of these rays.

(C) $X$-rays are electromagnetic waves with very short wavelengths,typically in the range of $0.01$ to $10$ nanometers.
Because they have very short wavelengths,they possess very high frequencies and high energy.
They are produced when high-energy electrons strike a metal target.
Unlike cathode rays (which are streams of electrons),$X$-rays do not carry any charge and are not deflected by electric or magnetic fields.
Therefore,$X$-rays are electromagnetic radiations of high frequency.
Thus,the correct option is $(c)$.

(C) The voltage applied across an $X$-ray tube is typically in the range of $10 \text{ kV}$ to $80 \text{ kV}$.
Since $1 \text{ kV} = 1000 \text{ V}$,the range is $10,000 \text{ V}$ to $80,000 \text{ V}$.
Among the given options,$10,000 \text{ V}$ is the most appropriate value.

(C) Characteristic $X-$ray radiation is produced due to electronic transitions within the atom. When high-energy bombarding electrons strike the target material,they can eject an electron from an inner shell (such as the $K$ or $L$ shell). This creates a vacancy in that shell. To restore stability,an electron from a higher energy outer shell drops down to fill this vacancy. The difference in energy between the two shells is released in the form of a photon,which corresponds to the characteristic $X-$ray radiation.

(B) In an $X-$ray tube,the target material must be a heavy element with a high melting point.
$1$. $A$ heavy element (high atomic number $Z$) is required to produce high-energy $X-$rays efficiently through the deceleration of electrons (Bremsstrahlung).
$2$. $A$ high melting point is essential because a significant portion of the kinetic energy of the incident electrons is converted into heat upon impact with the target. If the melting point were low,the target would melt under continuous operation.

(C) Moseley's law states that the frequency $\nu$ of a characteristic $X$-ray line is related to the atomic number $Z$ of the target element by the formula: $\nu = a(Z - b)^2$.
Here,$a$ and $b$ are constants that depend on the specific spectral line (e.g.,$K_{\alpha}, K_{\beta}$).
This law demonstrates that the frequency of characteristic $X$-rays is directly proportional to the square of the atomic number of the target material,rather than its atomic weight or density.
Therefore,the correct option is $(c)$.

(C) The Compton effect is the phenomenon of inelastic scattering of a photon by a charged particle,typically a free electron.
When a high-energy photon,such as an $X$-ray or a gamma-ray photon,collides with an electron,it transfers a portion of its energy to the electron.
This results in the photon being scattered at a specific angle with a lower energy and a longer wavelength compared to the incident photon.
Because this effect is most prominently observed and studied using $X$-rays,it is fundamentally associated with $X$-rays.

(B) $X-$rays and gamma rays are both high-energy electromagnetic waves. They share the same fundamental nature as they are part of the electromagnetic spectrum,travel at the speed of light,and do not carry any electric charge.

(C) The maximum energy of an $X$-ray photon is equal to the kinetic energy gained by an electron accelerated through a potential difference $V$.
$E_{\max} = eV = 1 \text{ } e \times 10^5 V = 10^5 \text{ } eV$.
To convert this into $MeV$,we divide by $10^6$:
$E_{\max} = \frac{10^5}{10^6} \text{ } MeV = 10^{-1} \text{ } MeV$.

(B) The energy of an $X$-ray photon is given by $E = h\nu = \frac{hc}{\lambda}$.
For the shortest wavelength (cutoff wavelength),the entire kinetic energy of the electron is converted into a single photon: $eV = \frac{hc}{\lambda_{\min}}$.
Rearranging for $\lambda_{\min}$,we get $\lambda_{\min} = \frac{hc}{eV}$.
Since $h$,$c$,and $e$ are constants,$\lambda_{\min} \propto \frac{1}{V}$.
Therefore,the shortest wavelength depends only on the accelerating voltage $V$ applied to the tube.

(C) The electromagnetic spectrum classifies $X-$rays as high-energy radiation with wavelengths typically ranging from $0.1 \ \mathring{A}$ to $100 \ \mathring{A}$.
Since $1 \ \mathring{A} = 10^{-10} \ m$,the order of magnitude for $X-$ray wavelengths is $10^{-10} \ m (\mathring{A}).$

(D) The production of $X$-rays is an atomic process,typically resulting from transitions of electrons between inner atomic shells.
Conversely,the production of $\gamma$-rays is a nuclear process,resulting from transitions between nuclear energy levels within the nucleus.
Therefore,$X$-rays and $\gamma$-rays of the same energy are distinguished by their method of production.

(A) The minimum wavelength (cut-off wavelength) of the continuous $X$-ray spectrum is given by the formula: $\lambda_{\min} = \frac{hc}{eV} = \frac{12400 \ \text{eV} \cdot \mathring{A}}{V \text{ (in volts)}}$.
Substituting the given potential difference $V = 40,000 \ V$:
$\lambda_{\min} = \frac{12400}{40000} \ \mathring{A} = 0.31 \ \mathring{A}$.
Since the $X$-ray tube cannot produce wavelengths shorter than the cut-off wavelength $\lambda_{\min}$,any wavelength $\lambda < 0.31 \ \mathring{A}$ will be absent from the spectrum.
Among the given options,$0.25 \ \mathring{A}$ is less than $0.31 \ \mathring{A}$.
Therefore,the wavelength $0.25 \ \mathring{A}$ is absent.

(A) 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 $E = eV = \frac{hc}{\lambda_{\min}}$.
Therefore,the minimum wavelength is given by $\lambda_{\min} = \frac{hc}{eV}$.
Since $E = eV$ is the energy of the electrons hitting the target,$\lambda_{\min}$ is inversely proportional to the energy of the electrons.

(B) The momentum $p$ of a photon is given by the de Broglie relation: $p = \frac{h}{\lambda}$.
Given:
Planck's constant $h = 6.626 \times 10^{-34} \text{ J s}$.
Wavelength $\lambda = 0.010 \mathring{A} = 0.010 \times 10^{-10} \text{ m} = 10^{-12} \text{ m}$.
Substituting the values:
$p = \frac{6.626 \times 10^{-34}}{10^{-12}} = 6.626 \times 10^{-22} \text{ kg m/s}$.

(A) Radar systems function by emitting electromagnetic waves that reflect off a target and return to the receiver to detect its position and speed.
$X$-rays have very high energy and short wavelengths,which causes them to be absorbed by most materials rather than being reflected.
Because they are not reflected by the target,they cannot be used for radar detection.
Therefore,the correct option is $A$.

(C) $X-$rays are high-energy electromagnetic waves that pass through soft tissues with very little absorption.
Since the intestines consist of soft tissues with similar densities,they do not absorb $X-$rays significantly.
Consequently,they do not cast a clear shadow on the photographic film,making it impossible to obtain a useful diagnostic image without the use of contrast agents like barium.

(D) $BaSO_4$ (Barium sulfate) contains Barium,which is a heavy element with a high atomic number $(Z = 56)$.
Because of its high atomic number and density,it is highly effective at absorbing $X$-rays.
When a patient drinks a $BaSO_4$ suspension,it coats the lining of the stomach.
Since $BaSO_4$ absorbs $X$-rays more strongly than the surrounding soft tissues,it creates a high-contrast image on the $X$-ray film,allowing doctors to clearly examine the structure of the stomach.

(A) To study the crystal structure,the wavelength of the incident radiation must be of the same order of magnitude as the interatomic spacing in the crystal lattice.
The interatomic spacing in crystals is typically in the range of $1 \,\mathring{A}$ to $3 \,\mathring{A}$.
$X$-rays have wavelengths in the range of $0.1 \,\mathring{A}$ to $10 \,\mathring{A}$.
Specifically,for diffraction by crystal planes (Bragg's Law),the wavelength should be comparable to the lattice constant,which is approximately $2 \,\mathring{A}$ to $0.1 \,\mathring{A}$.

(D) The emission of characteristic $X$-rays starts only when the incident electrons are accelerated beyond a certain critical value. Once this threshold is exceeded,the intensity of the characteristic lines relative to the continuous (white) spectrum increases,but the wavelengths of these characteristic lines remain unchanged because they depend only on the atomic structure of the target material.

(C) In an $X-$ray tube,electrons are accelerated by a potential difference $V$ and strike a metal target.
This process produces a continuous spectrum of $X-$rays due to Bremsstrahlung (braking radiation).
As an electron decelerates,it emits a photon with energy $E = hf = hc/\lambda$.
The maximum energy of the emitted photon corresponds to the kinetic energy of the incident electron,$E_{max} = eV = hc/\lambda_{min}$.
Therefore,the emitted $X-$ray spectrum contains all wavelengths greater than or equal to this minimum wavelength $\lambda_{min} = hc/eV$.
Thus,the beam consists of all wavelengths larger than a certain minimum wavelength.

(B) The continuous spectrum of $X-$rays is produced due to the deceleration of electrons as they strike the target material (Bremsstrahlung radiation).
Since the kinetic energy of an electron is $K = eV$, where $V$ is the accelerating voltage, the maximum energy of a photon emitted is equal to the kinetic energy of the electron.
Thus, $h \nu_{max} = eV$, which implies $hc / \lambda_{min} = eV$.
Therefore, $\lambda_{min} = hc / eV$.
This shows that there is a definite short wavelength limit (minimum wavelength) for a given accelerating voltage, while all wavelengths greater than this are present in the spectrum.

(B) The energy of an $X-$ray photon is given by the equation $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the radiation.
Penetrating power is directly proportional to the energy of the $X-$ray photons.
Since $E \propto \nu$,as the frequency $\nu$ increases,the energy of the photons increases.
Consequently,the penetrating power of the $X-$rays increases with an increase in their frequency.

(C) The electromagnetic spectrum is ordered by wavelength and frequency. Gamma rays have the highest frequency and the shortest wavelength in the electromagnetic spectrum.
$X$-rays have a longer wavelength compared to gamma rays but a shorter wavelength than ultraviolet rays.
Therefore,the wavelength of $X$-rays $({\lambda _1})$ is greater than the wavelength of gamma rays $({\lambda _2})$.
Thus,the correct relation is ${\lambda _1} > {\lambda _2}$.

(C) According to Moseley's law,the frequency $\nu$ of characteristic $X$-rays is given by $\nu = a(Z - b)^2$,where $a$ and $b$ are constants.
Since the frequency $\nu$ is related to the wavelength $\lambda$ by the relation $\nu = \frac{c}{\lambda}$,we have $\frac{c}{\lambda} \propto (Z - b)^2$.
For large values of $Z$,the constant $b$ can be neglected,leading to $\frac{1}{\lambda} \propto Z^2$.
Therefore,the wavelength $\lambda$ varies with the atomic number $Z$ as $\lambda \propto \frac{1}{Z^2}$.

(B) The energy of an $X$-ray photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
In an $X$-ray tube,the maximum energy of a photon is produced when an electron is completely stopped by the target,converting its kinetic energy into a photon.
The kinetic energy gained by an electron accelerated through a potential difference $V$ is $K = eV$,where $e$ is the elementary charge.
Equating the photon energy to the kinetic energy of the electron,we get $h\nu = eV$.
Since $h$ and $e$ are constants,it follows that $\nu \propto V$.

(C) The kinetic energy $K$ gained by an electron accelerated through a potential difference $V$ is given by $K = eV$.
In the production of $X$-rays,the maximum energy of a photon emitted during the collision of an electron with the target is equal to the kinetic energy of the electron.
Thus,$E_{\max} = h\nu_{\max} = eV$.
Solving for the maximum frequency $\nu_{\max}$,we get $\nu_{\max} = \frac{eV}{h}$.

(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K = eV$.
When this electron strikes the target,it produces $X$-rays. The maximum energy of the emitted $X$-ray photon corresponds to the case where the entire kinetic energy of the electron is converted into a single photon.
Thus,$E_{max} = h\nu_{max} = \frac{hc}{\lambda_{min}} = eV$.
Rearranging this equation for the minimum wavelength,we get $\lambda_{min} = \frac{hc}{eV}$.

(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula:
${\lambda _{\min }} = \frac{{hc}}{{eV}}$
where $h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $V$ is the accelerating potential difference.
From the relation ${\lambda _{\min }} \propto \frac{1}{V}$,it is clear that the minimum wavelength is inversely proportional to the applied potential difference.
Therefore,when the potential difference $V$ is increased,the minimum wavelength ${\lambda _{\min }}$ decreases.
Thus,the correct option is $(d)$.

(A) The energy gained by an electron accelerated through a potential difference $V$ is given by $E = eV$.
In an $X-ray$ tube, this energy is converted into the maximum energy of the emitted photon, which is $E = h\nu_{max}$.
Equating the two, we get $h\nu_{max} = eV$.
Therefore, the maximum frequency $\nu_{max}$ is given by $\nu_{max} = \frac{eV}{h}$.
Substituting the given values:
$\nu_{max} = \frac{1.6 \times 10^{-19} \ J \times 42,000 \ V}{6.63 \times 10^{-34} \ J \cdot s}$.
$\nu_{max} \approx \frac{6.72 \times 10^{-15}}{6.63 \times 10^{-34}} \approx 1.01 \times 10^{19} \ Hz$.
Rounding to the nearest order of magnitude, the maximum frequency is $10^{19} \ Hz$.

(D) In $K-$electron capture,the nucleus of an atom captures an electron from the innermost $K-$shell.
This creates a vacancy in the $K-$shell.
An electron from a higher energy shell (like the $L$ or $M$ shell) then transitions to fill this vacancy.
As the electron drops to a lower energy state,the energy difference is released in the form of a characteristic $X-$ray photon.

(D) Photons are the quanta of electromagnetic radiation.
Electromagnetic waves consist of oscillating electric and magnetic fields,but they do not carry any net electric charge or magnetic moment.
Since $X-$rays are electromagnetic waves,their photons are electrically neutral and possess no magnetic moment.
Therefore,the correct option is $(d)$.

(C) Hard $X$-rays are defined as those having high frequency and high penetrating power.
Since the frequency $\nu$ and wavelength $\lambda$ are related by the equation $c = \nu \lambda$,it follows that $\nu = \frac{c}{\lambda}$.
Therefore,a higher frequency corresponds to a shorter wavelength.
Among the given options,$0.1 \ \mathring{A}$ is the shortest wavelength.
Thus,$X$-rays with a wavelength of $0.1 \ \mathring{A}$ are the hardest.

(D) $X-$rays are electromagnetic waves consisting of oscillating electric and magnetic fields. Because they carry no net electric charge,they do not experience a Lorentz force when passing through external electric or magnetic fields. Therefore,$X-$ray beams remain unaffected by both electric and magnetic fields.

(B) $X-$rays are high-energy electromagnetic radiations produced when high-speed electrons strike a metal target.
These electrons interact with the inner-shell electrons of the target atoms.
When an inner-shell electron (e.g.,from the $K$ or $L$ shell) is knocked out,an electron from a higher energy level drops down to fill the vacancy.
The energy difference between these atomic energy levels is released in the form of a high-energy photon,which is an $X-$ray photon.
Therefore,$X-$rays are produced due to transitions between atomic energy levels.
Mathematically,the energy of the emitted $X-$ray is given by $\Delta E = E_{initial} - E_{final}$,and its wavelength is $\lambda = \frac{hc}{\Delta E}$.

(C) The electromagnetic spectrum is arranged in order of increasing frequency and decreasing wavelength. The order is: Radio waves,Microwaves,Infrared,Visible,Ultraviolet,$X$-rays,and Gamma rays. Therefore,the $X$-rays region lies between the ultraviolet region and the gamma rays region.

(B) The correct answer is $X-$ rays.
$X-$ rays have wavelengths comparable to the interatomic spacing in crystals (approximately $0.1 \ nm$ to $10 \ nm$).
Due to this,they undergo diffraction when passing through a crystal lattice.
This phenomenon,known as $X-$ ray diffraction,allows scientists to determine the internal atomic structure and arrangement of solid crystals.

(A) The intensity of an $X-ray$ beam is directly proportional to the number of electrons striking the target per unit time.
This number of electrons depends on the rate of thermionic emission from the filament.
The rate of thermionic emission is controlled by the filament current.
Therefore,increasing the filament current increases the number of electrons emitted,which in turn increases the intensity of the emitted $X-ray$ beam.
Thus,the correct option is $A$.

(C) To produce characteristic $X$-rays,the incident electron must have enough energy to knock out an electron from the inner shell of the target atom.
This requires the kinetic energy of the incident electron to be at least equal to the binding energy of the electron in that shell.
The kinetic energy of an electron accelerated through a potential difference $V$ is given by $K = eV$.
Therefore,to eject an electron from the innermost shell (binding energy $E_b = 40 \text{ keV}$),the condition is $eV > E_b$.
Substituting the values,$eV > 40 \text{ keV}$,which implies $V > 40 \text{ kV}$.

(A) The characteristic $X$-rays are produced when an electron from a higher energy shell (e.g.,$L, M, N, \dots$ shells) transitions to a vacancy in the innermost shell (the $K$-shell).
The energy of the emitted characteristic $X$-ray photon is given by the difference in energy between the two shells involved in the transition: $E_{X-ray} = E_{higher} - E_{lower}$.
Since the electron transitions from a higher energy level (closer to $0 \, eV$) to the innermost shell (which has a binding energy of $40 \, keV$,meaning its energy level is $-40 \, keV$),the energy released must be less than the binding energy of the innermost shell.
Therefore,the energy of the characteristic $X$-rays is always less than the binding energy of the innermost electron.

(D) The most energetic $X$-ray photon corresponds to the case where the entire kinetic energy of the electron is converted into a single photon.
$E = K.E. = 40 \ keV = 40 \times 10^3 \ eV$.
The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Thus,$\lambda = \frac{hc}{E}$.
Using the relation $\lambda (\text{in } \mathring{A}) \approx \frac{12400}{E (\text{in } eV)}$,we get:
$\lambda = \frac{12400}{40000} \mathring{A} = 0.31 \mathring{A}$.
Alternatively,using constants: $\lambda = \frac{(6.62 \times 10^{-34} \ J \cdot s) \times (3 \times 10^8 \ m/s)}{40 \times 10^3 \times 1.6 \times 10^{-19} \ J} = 0.309 \times 10^{-10} \ m = 0.309 \ \mathring{A} \approx 0.31 \ \mathring{A}$.

(C) $X$-rays are classified based on their penetrating power.
$X$-rays with higher energy and higher frequency have greater penetrating power and are known as hard $X$-rays.
Conversely,$X$-rays with lower energy and lower frequency have less penetrating power and are known as soft $X$-rays.
Therefore,$X$-rays that can penetrate through longer distances in substances are called hard $X$-rays.