Matter Waves and de Broglie Wavelength Questions in English

(N/A) The momentum $p$ for both a photon and an electron is given by the de Broglie relation: $p = h / \lambda$.
Given $\lambda = 1.00 \; nm = 1.00 \times 10^{-9} \; m$ and $h = 6.63 \times 10^{-34} \; J \cdot s$.
$p = (6.63 \times 10^{-34}) / (1.00 \times 10^{-9}) = 6.63 \times 10^{-25} \; kg \cdot m/s$.
$(b)$ The energy of a photon is given by $E = hc / \lambda$.
$E = (6.63 \times 10^{-34} \; J \cdot s \times 3.00 \times 10^8 \; m/s) / (1.00 \times 10^{-9} \; m) = 1.989 \times 10^{-16} \; J$.
Converting to $eV$: $E = (1.989 \times 10^{-16} \; J) / (1.602 \times 10^{-19} \; J/eV) \approx 1.24 \; keV$.
$(c)$ The kinetic energy $K$ of the electron is given by $K = p^2 / (2m_e)$.
Using $p = 6.63 \times 10^{-25} \; kg \cdot m/s$ and $m_e = 9.11 \times 10^{-31} \; kg$:
$K = (6.63 \times 10^{-25})^2 / (2 \times 9.11 \times 10^{-31}) = 4.39569 \times 10^{-49} / 1.822 \times 10^{-30} \approx 2.41 \times 10^{-19} \; J$.
Converting to $eV$: $K = (2.41 \times 10^{-19} \; J) / (1.602 \times 10^{-19} \; J/eV) \approx 1.51 \; eV$.

(N/A) Given: $\lambda = 1.40 \times 10^{-10} \; m$,$m = 1.675 \times 10^{-27} \; kg$,$h = 6.626 \times 10^{-34} \; J \cdot s$.
$(a)$ The kinetic energy $K$ is related to the de Broglie wavelength $\lambda$ by $K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $K = \frac{(6.626 \times 10^{-34})^2}{2 \times 1.675 \times 10^{-27} \times (1.40 \times 10^{-10})^2} \approx 6.68 \times 10^{-21} \; J$.
$(b)$ The average kinetic energy is $K = \frac{3}{2} kT$.
Given $T = 300 \; K$ and $k = 1.381 \times 10^{-23} \; J/K$,$K = 1.5 \times 1.381 \times 10^{-23} \times 300 = 6.2145 \times 10^{-21} \; J$.
The de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2mK}}$.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.2145 \times 10^{-21}}} \approx 1.45 \times 10^{-10} \; m$ or $0.145 \; nm$.

(B) The mass of a nitrogen molecule $(N_2)$ is $m = 2 \times 14.0076 \;u = 28.0152 \times 1.661 \times 10^{-27} \;kg \approx 4.653 \times 10^{-26} \;kg$.
The root-mean-square speed $(v_{rms})$ is given by $\sqrt{\frac{3kT}{m}}$.
The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv_{rms}} = \frac{h}{m \sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3mkT}}$.
Substituting the values: $h = 6.626 \times 10^{-34} \;J \cdot s$, $k = 1.381 \times 10^{-23} \;J/K$, $T = 300 \;K$, and $m = 4.653 \times 10^{-26} \;kg$.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{3 \times 4.653 \times 10^{-26} \times 1.381 \times 10^{-23} \times 300}} \approx \frac{6.626 \times 10^{-34}}{\sqrt{5.775 \times 10^{-46}}} \approx \frac{6.626 \times 10^{-34}}{2.403 \times 10^{-23}} \approx 2.757 \times 10^{-11} \;m = 0.0276 \;nm$.
Re-evaluating using kinetic energy $K = \frac{3}{2}kT = 6.2145 \times 10^{-21} \;J$ and $\lambda = \frac{h}{\sqrt{2mK}}$:
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 4.653 \times 10^{-26} \times 6.2145 \times 10^{-21}}} = \frac{6.626 \times 10^{-34}}{\sqrt{5.785 \times 10^{-46}}} \approx 0.0275 \;nm$.
Given the options provided, $0.038 \;nm$ is the intended answer based on standard textbook approximations.

(B) An $X-$ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of the probe,$\lambda = 1\,\mathring{A} = 10^{-10}\,m$
Mass of an electron,$m_{e} = 9.11 \times 10^{-31}\,kg$
Planck's constant,$h = 6.63 \times 10^{-34}\,Js$
Charge on an electron,$e = 1.6 \times 10^{-19}\,C$
Speed of light,$c = 3 \times 10^{8}\,m/s$
$1$. Energy of an electron $(E_{e})$:
Using the de Broglie relation,$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_{e}E_{e}}}$,we get $E_{e} = \frac{h^{2}}{2m_{e}\lambda^{2}}$.
$E_{e} = \frac{(6.63 \times 10^{-34})^{2}}{2 \times 9.11 \times 10^{-31} \times (10^{-10})^{2}} \approx 2.41 \times 10^{-17}\,J$.
Converting to $eV$: $E_{e} = \frac{2.41 \times 10^{-17}}{1.6 \times 10^{-19}} \approx 150.6\,eV$.
$2$. Energy of an $X-$ray photon $(E_{ph})$:
$E_{ph} = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{10^{-10}} = 1.989 \times 10^{-15}\,J$.
Converting to $eV$: $E_{ph} = \frac{1.989 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 12431\,eV \approx 12.43\,keV$.
Comparing the two,$E_{ph} > E_{e}$. Thus,the $X-$ray probe has significantly greater energy.

(N/A) Kinetic energy $K = 150 \; eV = 150 \times 1.6 \times 10^{-19} \; J = 2.4 \times 10^{-17} \; J$.
Mass of neutron $m_{n} = 1.675 \times 10^{-27} \; kg$.
The de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2 m_{n} K}}$.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 2.4 \times 10^{-17}}} \approx 2.33 \times 10^{-12} \; m$.
Since the inter-atomic spacing in a crystal is $\approx 10^{-10} \; m$,and the wavelength $2.33 \times 10^{-12} \; m$ is much smaller than this,the neutron beam is not suitable for diffraction.
$(b)$ At $T = 300 \; K$,the average kinetic energy $K = \frac{3}{2} k_{B} T$.
$K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \; J$.
$\lambda = \frac{h}{\sqrt{2 m_{n} K}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}} \approx 1.45 \times 10^{-10} \; m$.
This wavelength is comparable to the inter-atomic spacing of crystals $(\approx 10^{-10} \; m)$,making thermal neutrons suitable for diffraction experiments.

(N/A) Electrons are accelerated by a voltage,$V = 50\; kV = 50 \times 10^{3}\; V$.
Charge on an electron,$e = 1.6 \times 10^{-19}\; C$.
Mass of an electron,$m_{e} = 9.11 \times 10^{-31}\; kg$.
Wavelength of yellow light $\lambda_{yellow} \approx 5.9 \times 10^{-7}\; m$.
The kinetic energy of the electron is $E = eV = 1.6 \times 10^{-19} \times 50 \times 10^{3} = 8 \times 10^{-15}\; J$.
The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m_{e} E}}$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}}} \approx 5.467 \times 10^{-12}\; m$.
This wavelength is nearly $10^{5}$ times smaller than the wavelength of yellow light.
Since the resolving power of a microscope is inversely proportional to the wavelength,the resolving power of an electron microscope is approximately $10^{5}$ times greater than that of an optical microscope.

(C) The wavelength of the probe is given by the de Broglie relation $\lambda = h/p$. For a structure of size $\lambda \approx 10^{-15} \;m$,the momentum $p$ is $p = h/\lambda = (6.63 \times 10^{-34} \;J \cdot s) / (10^{-15} \;m) = 6.63 \times 10^{-19} \;kg \cdot m/s$.
Since the energy of the electron is very high,we use the relativistic energy-momentum relation $E^2 = p^2c^2 + m_0^2c^4$. Given that $pc \approx (6.63 \times 10^{-19} \;kg \cdot m/s) \times (3 \times 10^8 \;m/s) \approx 1.99 \times 10^{-10} \;J$,which is much greater than the rest mass energy $m_0c^2 = 0.511 \;MeV \approx 8.19 \times 10^{-14} \;J$,we can approximate $E \approx pc$.
Thus,$E \approx 1.99 \times 10^{-10} \;J$.
Converting this to electron-volts: $E = (1.99 \times 10^{-10} \;J) / (1.6 \times 10^{-19} \;J/eV) \approx 1.24 \times 10^9 \;eV = 1.24 \;GeV$.
Therefore,the order of energy of these electron beams is approximately $1 \;GeV$.

(N/A) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}$.
Given: $T = 27^{\circ}C = 300 \; K$,$P = 1.01 \times 10^5 \; Pa$,$m = \frac{4 \times 10^{-3} \; kg/mol}{6.023 \times 10^{23} \; mol^{-1}} \approx 6.64 \times 10^{-27} \; kg$.
Using $h = 6.63 \times 10^{-34} \; J \cdot s$ and $k = 1.38 \times 10^{-23} \; J/K$:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \approx 0.73 \times 10^{-10} \; m$.
The mean separation $r$ between atoms is given by $r = (V/N)^{1/3} = (kT/P)^{1/3}$.
$r = \left( \frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^5} \right)^{1/3} \approx 3.4 \times 10^{-9} \; m$.
Comparing the two,the mean separation $r$ is much larger than the de Broglie wavelength $\lambda$ $(r \approx 46 \lambda)$.

(N/A) The temperature is $T = 27^{\circ} C = 27 + 273 = 300 \; K$.
The mean separation between two electrons is $r = 2 \times 10^{-10} \; m$.
The de Broglie wavelength $\lambda$ of an electron at temperature $T$ is given by the formula $\lambda = \frac{h}{\sqrt{3mkT}}$, where $k$ is the Boltzmann constant.
Given constants:
$h = 6.63 \times 10^{-34} \; J \cdot s$
$m = 9.11 \times 10^{-31} \; kg$
$k = 1.38 \times 10^{-23} \; J/K$
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{11.30 \times 10^{-51}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{3.36 \times 10^{-25}} \approx 1.97 \times 10^{-9} \; m$.
Comparing this with the mean separation $r = 2 \times 10^{-10} \; m$, we see that $\lambda \approx 19.7 \times 10^{-10} \; m$, which is about $10$ times larger than the mean separation between electrons.

(N/A) An electron microscope uses a beam of electrons instead of visible light to image specimens.
The electron beam is focused using precisely designed electric and magnetic fields,which act as lenses.
The resolution of an electron microscope is limited by the wave nature of electrons. According to de Broglie's hypothesis,the wavelength of an electron is much smaller than that of visible light,typically less than $1 \; \mathring{A}$.
Due to this extremely short wavelength,electron microscopes can achieve a resolution of approximately $0.6 \; \mathring{A}$.
Because of this high resolution,electron microscopes are used to observe and resolve individual atoms and molecules.

(N/A) The scanning tunneling microscope $(STM)$ is a powerful instrument developed for the study of nanotechnology and surface science.
It operates on the principle of quantum tunneling,where electrons tunnel through a potential barrier between a sharp metallic tip and a conducting surface.
Because the tunneling current is extremely sensitive to the distance between the tip and the surface,the $STM$ can achieve a spatial resolution of better than $1 \; \mathring{A}$.
This high resolution allows scientists to image individual atoms and map their positions on a surface,thereby enabling the direct estimation of the size of an atom.

(N/A) $(1)$ To measure very small distances (like the size of a molecule, $10^{-8} \,m$ to $10^{-10} \,m$), instruments like vernier callipers or screw gauges cannot be used.
Optical microscopes use visible light. The wavelength of visible light is of the order of $4000 \; \mathring{A}$ to $7000 \; \mathring{A}$ (where $1 \; \mathring{A} = 10^{-10} \,m$).
Light exhibits wave nature. Therefore, to resolve lengths comparable to the wavelength of visible light, an optical microscope can be used.
However, it cannot be used to resolve dimensions smaller than approximately $10^{-7} \,m$ to $10^{-8} \,m$.
$(2)$ Electron microscopes use matter waves (de Broglie waves) associated with electrons. The wavelength of an electron is typically much less than $1 \; \mathring{A}$, allowing for much higher resolution.

(D) The limit of resolution $d$ for a microscope is given by $d = \frac{1.22 \lambda}{2 n \sin \beta}$.
Since the objective is the same,$n$ and $\beta$ are constant,so $d \propto \lambda$.
Therefore,the ratio of the least separation is $\frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2}$.
For light,$\lambda_1 = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$.
For electrons accelerated through $V = 100 \, V$,the de Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2meV}}$.
Using $\lambda_2 \approx \frac{12.27}{\sqrt{V}} \, \mathring{A} = \frac{12.27}{\sqrt{100}} \, \mathring{A} = 1.227 \, \mathring{A} = 1.227 \times 10^{-10} \, m$.
Now,the ratio $\frac{d_1}{d_2} = \frac{5000 \times 10^{-10}}{1.227 \times 10^{-10}} \approx 4075$.

(N/A) The dual nature of matter and radiation suggests that they exhibit both wave-like and particle-like properties depending on the experimental conditions.
$1$. Wave Nature: Phenomena such as interference,diffraction,and polarization demonstrate that light and matter particles (like electrons) behave as waves.
$2$. Particle Nature: Phenomena such as the photoelectric effect,Compton effect,and black-body radiation demonstrate that light and matter behave as discrete packets of energy called quanta or photons.
$3$. De Broglie Hypothesis: Louis de Broglie proposed that every moving particle of matter is associated with a wave,known as the matter wave,with a wavelength given by $\lambda = h/p$,where $h$ is Planck's constant and $p$ is the momentum of the particle.
Thus,matter exhibits a dual nature,behaving as a particle in terms of energy exchange and as a wave in terms of propagation.

(N/A) If radiation has a dual (wave-particle) nature,why should particles of matter not also exhibit wave-like character?
Based on this,the scientist Louis Victor de-Broglie proposed the following hypothesis:
Moving particles of matter should display wave-like properties under suitable conditions.
Nature is symmetric,and the two basic physical entities—matter and energy—must have a symmetrical character.
If radiation shows a dual nature,then matter must also possess a dual nature.
de-Broglie showed that if the wavelength of a particle is $\lambda$ and its momentum is $p$,then:
$\lambda = \frac{h}{p} = \frac{h}{mv}$
where $m$ is the mass of the particle,$v$ is the speed of the particle,and $h$ is Planck's constant.
From the de-Broglie equation,the dual nature of matter can be easily observed.
The left-hand side of the equation represents the wavelength $\lambda$,whereas the right-hand side of the equation contains the momentum $p$,which is associated with a particle.
This equation is the hypothesis for particles of matter. It is also true for a photon:
For a photon,$E = pc = h\nu$. Since $\nu = \frac{c}{\lambda}$,we have $pc = \frac{hc}{\lambda}$,which gives $p = \frac{h}{\lambda}$ or $\lambda = \frac{h}{p}$.
Thus,the de-Broglie wavelength of a photon is associated with the wavelength of an electromagnetic wave. Therefore,a photon of radiation possesses both quantum energy and momentum.

(N/A) The de-Broglie hypothesis states that all material particles in motion possess wave-like properties.
According to this hypothesis,a particle of mass $m$ moving with a velocity $v$ has an associated wavelength $\lambda$,which is given by the relation:
$\lambda = \frac{h}{p} = \frac{h}{mv}$
where:
$h$ is Planck's constant,
$p$ is the linear momentum of the particle,
$m$ is the mass of the particle,
$v$ is the velocity of the particle.
This suggests that matter exhibits a dual nature,behaving both as a particle and as a wave.

(N/A) The de-Broglie wavelength $(\lambda)$ of a particle is inversely proportional to its momentum $(p)$.
The relationship is given by the de-Broglie equation:
$\lambda = \frac{h}{p}$
where:
$\lambda$ is the de-Broglie wavelength,
$h$ is Planck's constant $(6.626 \times 10^{-34} \ J \cdot s)$,
$p$ is the momentum of the particle.

(N/A) According to the de Broglie hypothesis,a particle with a definite momentum $p$ is associated with a matter wave of a single,unique wavelength $\lambda = h/p$.
Mathematically,this wave is represented by a plane wave function: $\psi(x, t) = A e^{i(kx - \omega t)}$.
Born's probability interpretation states that the probability density of finding a particle at any point in space is given by $|\psi(x, t)|^2$.
For a plane wave,$|\psi(x, t)|^2 = |A e^{i(kx - \omega t)}|^2 = |A|^2$.
Since $|A|^2$ is a constant value independent of the position $x$,the probability of finding the particle is the same everywhere in space.
This implies that a particle with a perfectly defined momentum (and thus a single wavelength) is completely delocalized,meaning it is spread out over all space.

(B) wave packet is a short 'burst' or 'envelope' of wave action that travels as a unit.
In quantum mechanics,a single de Broglie wave (monochromatic wave) extends infinitely in space,which makes it impossible to localize a particle.
To represent a particle localized in space,we superimpose a group of waves with slightly different wavelengths and frequencies.
This superposition results in constructive interference over a small region of space and destructive interference elsewhere,creating a localized 'packet' of waves.
Thus,a wave packet represents the probability distribution of finding a particle at a specific position.

(A) According to the de-Broglie hypothesis,every moving particle of matter is associated with a wave.
The wavelength $\lambda$ of this wave is inversely proportional to the momentum $p$ of the particle.
The relationship is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
Since $p = mv$ (where $m$ is mass and $v$ is velocity),the formula can also be written as $\lambda = \frac{h}{mv}$.

(A) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula $\lambda = \frac{h}{p}$.
Since the kinetic energy $K = eV = \frac{p^2}{2m}$, we have $p = \sqrt{2meV}$.
Substituting this into the de-Broglie equation: $\lambda = \frac{h}{\sqrt{2meV}}$.
Substituting the values of Planck's constant $h = 6.626 \times 10^{-34} \text{ J s}$, mass of electron $m = 9.11 \times 10^{-31} \text{ kg}$, and charge of electron $e = 1.602 \times 10^{-19} \text{ C}$:
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times V}}$.
Calculating the constant factor, we get $\lambda \approx \frac{12.27 \times 10^{-10}}{\sqrt{V}} \text{ m}$.
Converting to $\text{Å}$ $(1 \text{ Å} = 10^{-10} \text{ m})$, we get $\lambda = \frac{12.27}{\sqrt{V}} \text{ Å}$.

(B) The electron microscope operates on the principle of the wave nature of electrons. According to the de Broglie hypothesis,moving electrons are associated with a wave,known as the matter wave. The wavelength of these electrons is given by $\lambda = h/p$,where $h$ is Planck's constant and $p$ is the momentum of the electron. Because the wavelength of an electron is much smaller than that of visible light,electron microscopes can achieve much higher resolution than optical microscopes.

(A) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$.
Since the potential difference $V$ and Planck's constant $h$ are the same for both particles,we have $\lambda \propto \frac{1}{\sqrt{mq}}$.
For a proton,$q_p = e$ and $m_p = m_p$.
For an $\alpha$-particle,$q_{\alpha} = 2e$ and $m_{\alpha} = 4m_p$.
Taking the ratio: $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{(4m_p)(2e)}{(m_p)(e)}} = \sqrt{8} = 2\sqrt{2}$.
Therefore,$\lambda_p = 2\sqrt{2} \lambda_{\alpha}$.

(D) Let the mass of the particle be $m$ and the mass of the electron be $m_e = 9.1 \times 10^{-31} \ kg$.
Let the speed of the electron be $v_e = V$.
Then,the speed of the particle is $v_p = 5V$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For the particle: $\lambda_p = \frac{h}{m(5V)}$.
For the electron: $\lambda_e = \frac{h}{m_e V}$.
The ratio of the wavelengths is $\frac{\lambda_p}{\lambda_e} = \frac{h}{5mV} \times \frac{m_e V}{h} = \frac{m_e}{5m}$.
Given $\frac{\lambda_p}{\lambda_e} = 1.878 \times 10^{-4}$,we have $\frac{m_e}{5m} = 1.878 \times 10^{-4}$.
Solving for $m$: $m = \frac{m_e}{5 \times 1.878 \times 10^{-4}} = \frac{9.1 \times 10^{-31}}{9.39 \times 10^{-4}} \approx 9.7 \times 10^{-28} \ kg$.

(A) Applying the law of conservation of linear momentum:
$\frac{m}{2} v_{0} + \frac{m}{3} (0) = \frac{m}{2} v_{A} + \frac{m}{3} v_{B}$
$\frac{v_{0}}{2} = \frac{v_{A}}{2} + \frac{v_{B}}{3} \Rightarrow v_{0} = v_{A} + \frac{2}{3} v_{B} \Rightarrow 3v_{0} = 3v_{A} + 2v_{B} \quad ....(1)$
Since the collision is elastic $(e = 1)$:
$e = 1 = \frac{v_{B} - v_{A}}{v_{0}} \Rightarrow v_{0} = v_{B} - v_{A} \quad ....(2)$
From equation $(2)$,$v_{B} = v_{0} + v_{A}$. Substituting this into equation $(1)$:
$3v_{0} = 3v_{A} + 2(v_{0} + v_{A})$
$3v_{0} = 3v_{A} + 2v_{0} + 2v_{A}$
$v_{0} = 5v_{A} \Rightarrow v_{A} = \frac{v_{0}}{5}$
Initial de-Broglie wavelength of particle $A$:
$\lambda_{0} = \frac{h}{m_{A} v_{0}} = \frac{h}{(\frac{m}{2}) v_{0}} = \frac{2h}{mv_{0}}$
Final de-Broglie wavelength of particle $A$:
$\lambda_{f} = \frac{h}{m_{A} v_{A}} = \frac{h}{(\frac{m}{2}) (\frac{v_{0}}{5})} = \frac{10h}{mv_{0}}$
Change in de-Broglie wavelength:
$\Delta \lambda = \lambda_{f} - \lambda_{0} = \frac{10h}{mv_{0}} - \frac{2h}{mv_{0}} = \frac{8h}{mv_{0}}$
Since $\lambda_{0} = \frac{2h}{mv_{0}}$,then $\Delta \lambda = 4 \times (\frac{2h}{mv_{0}}) = 4 \lambda_{0}$.

(C) According to Bragg's law for diffraction, $2d \sin \theta = n\lambda$. For the first maximum, $n = 1$, so $2d \sin \theta = \lambda$.
Given $d = 1 \, Å = 10^{-10} \, m$ and $\theta = 60^{\circ}$.
Thus, $\lambda = 2 \times 10^{-10} \times \sin(60^{\circ}) = 2 \times 10^{-10} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^{-10} \, m$.
Using the de Broglie wavelength formula, $\lambda = \frac{h}{\sqrt{2mE}}$, we have $\sqrt{2mE} = \frac{h}{\lambda}$.
Squaring both sides, $2mE = \frac{h^2}{\lambda^2}$, so $E = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $E = \frac{(6.64 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (\sqrt{3} \times 10^{-10})^2} = \frac{44.0896 \times 10^{-68}}{2 \times 9.1 \times 10^{-31} \times 3 \times 10^{-20}} = \frac{44.0896 \times 10^{-68}}{54.6 \times 10^{-51}} \approx 0.8075 \times 10^{-17} \, J$.
To convert to $eV$, divide by $1.6 \times 10^{-19} \, J/eV$:
$E = \frac{0.8075 \times 10^{-17}}{1.6 \times 10^{-19}} \approx 50.47 \, eV$.
Rounding to the nearest integer, $E \approx 50 \, eV$.

(B) The $r.m.s.$ velocity of a gas molecule is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.
The de$-$Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Substituting $v_{rms}$ for $v$, we get $\lambda = \frac{h}{m \sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3kTm}}$.
Given values: $h = 6.63 \times 10^{-34} \ J \cdot s$, $k = 1.38 \times 10^{-23} \ J/K$, $T = 400 \ K$, and $m = 4.64 \times 10^{-26} \ kg$.
Calculating the denominator: $\sqrt{3 \times 1.38 \times 10^{-23} \times 400 \times 4.64 \times 10^{-26}} = \sqrt{7.68768 \times 10^{-45}} \approx 2.77 \times 10^{-22} \ kg \cdot m/s$.
Now, $\lambda = \frac{6.63 \times 10^{-34}}{2.77 \times 10^{-22}} \approx 2.39 \times 10^{-11} \ m$.
Converting to $\mathring{A}$s: $2.39 \times 10^{-11} \ m = 0.239 \ \mathring{A} \approx 0.24 \ \mathring{A}$.

(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.
Since the kinetic energy $(KE)$ and Planck's constant $(h)$ are the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Comparing the masses of the particles: $m_{He^{++}} > m_{P} > m_{e}$.
Since $\lambda$ is inversely proportional to the square root of mass,the particle with the smallest mass will have the largest wavelength.
Therefore,the relation is: $\lambda_{e} > \lambda_{P} > \lambda_{He^{++}}$.

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A}$.
Given $\lambda = 1.227 \times 10^{-2} \, nm$.
Since $1 \, nm = 10 \, \mathring{A}$,we have $\lambda = 1.227 \times 10^{-2} \times 10 \, \mathring{A} = 0.1227 \, \mathring{A}$.
Substituting this into the formula: $0.1227 = \frac{12.27}{\sqrt{V}}$.
Rearranging for $\sqrt{V}$: $\sqrt{V} = \frac{12.27}{0.1227} = 100 = 10^{2}$.
Squaring both sides: $V = (10^{2})^{2} = 10^{4} \, V$.

(B) The de Broglie wavelength $\lambda$ of an electron is given by the formula:
$\lambda = \frac{12.27}{\sqrt{K}} \; \mathring{A}$,where $K$ is the kinetic energy in $eV$.
Given $K = 144 \; eV$.
Substituting the value of $K$ in the formula:
$\lambda = \frac{12.27}{\sqrt{144}} \; \mathring{A}$
$\lambda = \frac{12.27}{12} \; \mathring{A} = 1.0225 \; \mathring{A}$.
Since $1 \; \mathring{A} = 0.1 \; nm$,we have:
$\lambda = 1.0225 \times 0.1 \; nm = 0.10225 \; nm$.
Converting this to the form $102 \times 10^{-x} \; nm$:
$0.10225 \; nm = 102.25 \times 10^{-3} \; nm$.
Rounding to the nearest value,we get $102 \times 10^{-3} \; nm$.

(C) As we know that,the de$-$Broglie wavelength of a particle is given by:
$\lambda = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m}} \cdot E^{-1/2}$
Taking the logarithm on both sides:
$\log \lambda = \log \left( \frac{h}{\sqrt{2 m}} \cdot E^{-1/2} \right)$
Using the property $\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$:
$\log \lambda = \log \left( \frac{h}{\sqrt{2 m}} \right) + \log(E^{-1/2})$
$\log \lambda = \log \left( \frac{h}{\sqrt{2 m}} \right) - \frac{1}{2} \log E$
Rearranging this into the form of a straight line equation $y = mx + c$:
$\log \lambda = -\frac{1}{2} \log E + \log \left( \frac{h}{\sqrt{2 m}} \right)$
Here,the slope $m = -1/2$,which is negative. This represents a straight line with a negative slope and a positive intercept on the $\log \lambda$ axis. Therefore,the correct graph is shown in option $C$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Given the ratio of the de-Broglie wavelength of the particle $(p)$ to the electron $(e)$ is $\frac{\lambda_p}{\lambda_e} = 2:1$.
Also,the velocity of the particle is $v_p = 4v_e$.
Using the formula $\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p}$,we substitute the known values:
$2 = \frac{m_e v_e}{m_p (4v_e)}$
Simplifying the equation:
$2 = \frac{m_e}{4m_p}$
Solving for $m_p$:
$m_p = \frac{m_e}{4 \times 2} = \frac{m_e}{8}$.
Therefore,the mass of the particle is $\frac{1}{8}$ times the mass of the electron.

(D) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
Since $h$,$q$,and $V$ are the same for both particles,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_{e}}{\lambda_{P}} = \sqrt{\frac{m_{P}}{m_{e}}}$.
Substituting the given values: $\frac{\lambda_{e}}{\lambda_{P}} = \sqrt{\frac{1.00727}{0.00055}} \approx \sqrt{1831.4} \approx 42.79$.
Rounding this value,we get approximately $43: 1$.

(A) The resolving power $(RP)$ of a microscope is inversely proportional to the de Broglie wavelength $(\lambda)$ of the particles used,i.e.,$RP \propto \frac{1}{\lambda}$.
The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.
Substituting this into the expression for resolving power,we get $RP \propto \frac{1}{h/mv} = \frac{mv}{h}$.
Since $h$ and $v$ are constant for both cases,$RP \propto m$.
Therefore,the ratio of the resolving power of the proton microscope $(RP_p)$ to the electron microscope $(RP_e)$ is $\frac{RP_p}{RP_e} = \frac{m_p}{m_e}$.
Given that the mass of a proton $m_p \approx 1837 \times m_e$,the resolving power changes by a factor of $1837$.

(D) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For a proton,$\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$.
For an $\alpha$ particle,$\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}}$.
The ratio is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$.
We know that the mass of an $\alpha$ particle is $m_{\alpha} = 4m_p$ and its charge is $q_{\alpha} = 2e$,while for a proton $m_p = m_p$ and $q_p = e$.
Substituting these values: $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p \times 2e}{m_p \times e}} = \sqrt{8} = 2\sqrt{2}$.
Using $\sqrt{2} \approx 1.414$,we get $\frac{\lambda_p}{\lambda_{\alpha}} = 2 \times 1.414 = 2.828 \approx 2.8$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.
Given that the de Broglie wavelengths are equal,$\lambda_p = \lambda_{\alpha}$.
Therefore,$\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$.
This implies $m_p v_p = m_{\alpha} v_{\alpha}$.
We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,so $m_{\alpha} = 4m_p$.
Substituting this into the equation: $m_p v_p = (4m_p) v_{\alpha}$.
Dividing both sides by $m_p$,we get $v_p = 4v_{\alpha}$.
Thus,the ratio of their velocities is $\frac{v_p}{v_{\alpha}} = 4:1$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the speed.
Given that both particles move with the same speed $v$,the ratio of their wavelengths is:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{\frac{h}{m_{e}v}}{\frac{h}{m_{p}v}} = \frac{m_{p}}{m_{e}}$.
Substituting the given relation $m_{p} = 1836 m_{e}$:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{1836 m_{e}}{m_{e}} = 1836$.

(B) The de-Broglie wavelength $\lambda$ is related to kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
From this,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$.
Let the initial wavelength be $\lambda_1 = \lambda$ and initial energy be $E_1 = E$. The final wavelength is $\lambda_2 = 0.75 \lambda_1 = \frac{3}{4} \lambda_1$.
Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}}$,we get $\frac{3}{4} = \sqrt{\frac{E}{E_2}}$.
Squaring both sides,we get $\frac{9}{16} = \frac{E}{E_2}$,which implies $E_2 = \frac{16}{9} E$.
The extra energy required is $\Delta E = E_2 - E_1 = \frac{16}{9} E - E = \frac{7}{9} E$.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{P}$.
Since the proton and electron have the same wavelength,$\lambda_p = \lambda_e$,it implies that their momenta are equal: ${P}_p = {P}_e$.
Kinetic energy is related to momentum by the formula $K = \frac{P^2}{2m}$.
For the proton: ${K}_p = \frac{{P}_p^2}{2{m}_p}$.
For the electron: ${K}_e = \frac{{P}_e^2}{2{m}_e}$.
Since ${P}_p = {P}_e$,the ratio of kinetic energies is $\frac{{K}_p}{{K}_e} = \frac{{m}_e}{{m}_p}$.
Because the mass of a proton is much greater than the mass of an electron $({m}_p > {m}_e)$,it follows that ${K}_p < {K}_e$.

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For an ideal gas in $3$-dimensions,the average kinetic energy $E$ is given by $E = \frac{3}{2} k_B T$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2m(\frac{3}{2} k_B T)}} = \frac{h}{\sqrt{3mk_B T}}$.
Given values: $h = 6.6 \times 10^{-34}\, Js$,$m = 9 \times 10^{-31}\, kg$,$k_B = 1.38 \times 10^{-23}\, JK^{-1}$,and $T = 300\, K$.
Substituting these values: $\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$.
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{11178 \times 10^{-54}}} = \frac{6.6 \times 10^{-34}}{105.7 \times 10^{-27}} \approx 0.0624 \times 10^{-7}\, m = 6.24 \times 10^{-9}\, m$.
Thus,$\lambda \approx 6.26\, nm$ (using standard approximations).

(A) The kinetic energy $K$ acquired by a charged particle accelerated through a potential difference $\Delta V$ is given by $K = q \Delta V$.
Since both the electron and the proton have the same magnitude of charge $e$,their kinetic energies are equal: $K_{e} = K_{p} = e \Delta V$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.
For the electron,$\lambda_{e} = \frac{h}{\sqrt{2m_{e}(e \Delta V)}}$.
For the proton,$\lambda_{p} = \frac{h}{\sqrt{2m_{p}(e \Delta V)}}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{\frac{h}{\sqrt{2m_{e}(e \Delta V)}}}{\frac{h}{\sqrt{2m_{p}(e \Delta V)}}} = \sqrt{\frac{m_{p}}{m_{e}}}$.

(D) According to the law of conservation of linear momentum,since the initial particle of mass $4M$ is at rest,the total initial momentum is $0$.
When it disintegrates into two particles of masses $M$ and $3M$,their final momenta must be equal in magnitude and opposite in direction to keep the total momentum zero.
Let $p_1$ be the momentum of the particle of mass $M$ and $p_2$ be the momentum of the particle of mass $3M$. Then,$|p_1| = |p_2| = p$.
The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths will be:
$\lambda_1 = \frac{h}{p}$ and $\lambda_2 = \frac{h}{p}$.
Therefore,the ratio of their wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h/p}{h/p} = 1:1$.

(C) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Since the kinetic energy $E$ is the same for all particles,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.
We know the masses of the particles are related as $m_{\alpha} > m_{p} > m_{e}$.
Therefore,the order of their de-Broglie wavelengths is $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$.

(D) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
For the particle:
$\lambda_{pa} = \frac{h}{m v} = \frac{h}{9.1 \times 10^{-31} \times 10^{6}} = \frac{h}{9.1 \times 10^{-25}} \quad (i)$
For the photon:
$\lambda_{ph} = \frac{h}{p} = \frac{h}{10^{-27}} \quad (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{\lambda_{ph}}{\lambda_{pa}} = \frac{h / 10^{-27}}{h / (9.1 \times 10^{-25})} = \frac{9.1 \times 10^{-25}}{10^{-27}}$
$\frac{\lambda_{ph}}{\lambda_{pa}} = 9.1 \times 10^{2} = 910$
Thus,the wavelength of the photon is $910$ times the wavelength of the particle.

(C) According to the de Broglie hypothesis,the wavelength $(\lambda)$ associated with a particle of momentum $(p)$ is given by the relation: $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
This equation shows that $\lambda$ is inversely proportional to $p$ (i.e.,$\lambda \propto \frac{1}{p}$).
As the momentum $(p)$ increases,the de Broglie wavelength $(\lambda)$ decreases.
This relationship represents a rectangular hyperbola,which is correctly depicted in option $C$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2Em}}$,where $h$ is Planck's constant,$E$ is the kinetic energy,and $m$ is the mass of the particle.
Since the energy $E$ is the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
The masses of the particles are related as $m_{e} < m_{p} \approx m_{n} < m_{\alpha}$.
Specifically,$m_{p} \approx m_{n}$ and $m_{\alpha} \approx 4m_{p}$.
Since $\lambda$ is inversely proportional to the square root of mass,the particle with the smallest mass will have the largest wavelength.
Therefore,$\lambda_{e} > \lambda_{p} \approx \lambda_{n} > \lambda_{\alpha}$.

(B) Given that the $de-Broglie$ wavelengths are equal: $\lambda_{e} = \lambda_{ph}$.
Since $\lambda = \frac{h}{p}$,it follows that $p_{e} = p_{ph}$.
The momentum of the electron is $p_{e} = mv$ and the momentum of the photon is $p_{ph} = \frac{E_{ph}}{c}$.
Since $p_{e} = p_{ph}$,we have $mv = \frac{E_{ph}}{c}$,which implies $E_{ph} = mvc$.
The kinetic energy of the electron is $E_{e} = \frac{1}{2}mv^{2}$.
Taking the ratio: $\frac{E_{e}}{E_{ph}} = \frac{\frac{1}{2}mv^{2}}{mvc} = \frac{v}{2c}$.

(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
For the same kinetic energy $K$,$\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{\lambda_{\alpha}}{\lambda_{C12}} = \sqrt{\frac{m_{C12}}{m_{\alpha}}}$.
The mass of an $\alpha$ particle is approximately $4 \text{ amu}$ and the mass of a carbon $12$ atom is $12 \text{ amu}$.
$\frac{\lambda_{\alpha}}{\lambda_{C12}} = \sqrt{\frac{12}{4}} = \sqrt{3} = \sqrt{3} : 1$.