De Broglie's principle Questions in English

(C) During the experimental verification of the de-Broglie hypothesis,Davisson and Germer confirmed the wave nature of the electron through electron diffraction experiments.

(C) According to the De Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Here,$h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.
For a particle to exhibit significant wave motion,its wavelength $\lambda$ must be large enough to be observable.
Since $\lambda$ is inversely proportional to mass $m$ and velocity $v$,the wavelength becomes significant only when the mass $m$ of the particle is extremely small (negligible),such as in the case of an electron.

(A) The momentum $(p)$ of a photon is given by the de Broglie relation: $p = \frac{h}{\lambda}$.
Given:
$h = 6.6 \times 10^{-34} \ J \ s$
$\lambda = 2.2 \times 10^{-11} \ m$
Substituting the values:
$p = \frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}} = 3 \times 10^{-23} \ kg \ m \ s^{-1}$.
Therefore,the correct option is $(A)$.

(C) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its velocity (speed).
From this relation,it is clear that $\lambda \propto \frac{1}{v}$.
Therefore,as the speed $(v)$ of the electron increases,the wavelength $(\lambda)$ associated with it decreases.

(C) The de Broglie relation is given by $\lambda = \frac{h}{mv}$,where $m$ is the mass of the photon,$h$ is Planck's constant $(6.626 \times 10^{-34} \ J \ s)$,and $v$ is the velocity of the photon.
Rearranging for mass: $m = \frac{h}{v \lambda}$.
The velocity of a photon $(v)$ is $3 \times 10^{8} \ m \ s^{-1}$.
The wavelength $(\lambda)$ is $1.54 \times 10^{-8} \ cm = 1.54 \times 10^{-10} \ m$.
Substituting the values: $m = \frac{6.626 \times 10^{-34} \ J \ s}{(3 \times 10^{8} \ m \ s^{-1}) \times (1.54 \times 10^{-10} \ m)}$.
$m = \frac{6.626 \times 10^{-34}}{4.62 \times 10^{-2}} \ kg = 1.434 \times 10^{-32} \ kg$.
Rounding to the nearest provided option,the correct answer is $1.4285 \times 10^{-32} \ kg$.

(C) The de Broglie equation is given by $\lambda = \frac{h}{mv} = \frac{h}{p}$,where $\lambda$ is the wavelength,$h$ is Planck's constant,$m$ is the mass,$v$ is the velocity,and $p$ is the momentum of the particle.
Thus,the wavelength is inversely proportional to the momentum $(p = mv)$ of the electron.

(A) Louis-Victor de Broglie proposed the wave-particle duality of matter,stating that electrons exhibit wave-like properties. This concept is known as the de Broglie hypothesis,for which he was awarded the Nobel Prize in Physics.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Since the momentum is defined as $p = mv$,we can substitute $p$ into the equation.
Therefore,$\lambda = \frac{h}{p}$.
This is known as the de Broglie relation.

(B) The de-Broglie equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$,relates the wavelength (wave property) to the momentum (particle property) of a moving object,including photons. This equation provides the mathematical basis for the wave-particle duality of light and matter.

(B) The de-Broglie relationship states that the wavelength $(\lambda)$ of a particle is inversely proportional to its momentum $(p = mv)$.
Mathematically,this is expressed as $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.
Therefore,the correct expression is $\lambda = \frac{h}{mv}$.

(D) The de-Broglie equation relates the wavelength $(\lambda)$ of a particle to its momentum $(p = mv)$.
It is given by the formula: $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.
Therefore,the correct option is $(D)$.

(A) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Given: mass $m = 1 \, g = 10^{-3} \, kg$,velocity $v = 100 \, m/sec$,and Planck's constant $h = 6.63 \times 10^{-34} \, J \cdot s$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34} \, J \cdot s}{(10^{-3} \, kg) \times (100 \, m/sec)} = \frac{6.63 \times 10^{-34}}{10^{-1}} = 6.63 \times 10^{-33} \, m$.
Therefore,the correct option is $A$.

(D) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Assuming the same velocity $v$ for all particles,the wavelength is inversely proportional to the mass,i.e.,$\lambda \propto \frac{1}{m}$.
Comparing the masses: $m(\text{electron}) < m(\text{proton}) < m(CO_2) < m(SO_2)$.
Since the $SO_2$ molecule has the highest molecular mass $(64 \ g/mol)$,it will have the minimum de-Broglie wavelength.
Therefore,the correct option is $D$.

(D) The de-Broglie wavelength $(\lambda)$ is related to the momentum $(p)$ of a particle by the equation: $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
From this equation,it is clear that $\lambda$ is inversely proportional to the momentum $(p)$.

(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.
Since kinetic energy $KE = \frac{p^2}{2m}$,we have $p = \sqrt{2mKE}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mKE}}$.
Given $h = 6.62 \times 10^{-34} \ J \cdot s$,$m = 9.1 \times 10^{-31} \ kg$,and $KE = 2.8 \times 10^{-23} \ J$.
$\lambda = \frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.8 \times 10^{-23}}}$.
$\lambda = \frac{6.62 \times 10^{-34}}{\sqrt{50.96 \times 10^{-54}}} = \frac{6.62 \times 10^{-34}}{7.138 \times 10^{-27}} \approx 9.28 \times 10^{-8} \ m$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \ J \ s$
$m = 9.1 \times 10^{-31} \ kg$
$v = 1.2 \times 10^5 \ m \ s^{-1}$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 1.2 \times 10^5}$
$\lambda = \frac{6.626 \times 10^{-34}}{10.92 \times 10^{-26}}$
$\lambda \approx 0.6068 \times 10^{-8} \ m = 6.068 \times 10^{-9} \ m$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given: Mass $(m) = 10^{-6} \ kg$,Velocity $(v) = 10 \ ms^{-1}$,and Planck's constant $(h) = 6.63 \times 10^{-34} \ J \cdot s$.
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{10^{-6} \times 10} = \frac{6.63 \times 10^{-34}}{10^{-5}} = 6.63 \times 10^{-29} \ m$.
Therefore,the correct option is $(B)$.

(B) The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{mv}$.
From Bohr's postulate,$mvr = \frac{nh}{2\pi}$,which implies $\frac{h}{mv} = \frac{2\pi r}{n}$.
Thus,$\lambda = \frac{2\pi r}{n}$.
For the $3^{rd}$ orbit of hydrogen,$n = 3$ and the radius $r = n^2 a_0 = 3^2 \times 0.529 \, \mathring{A} = 9 \times 0.529 \, \mathring{A}$.
Substituting these values: $\lambda = \frac{2\pi \times 9 \times 0.529}{3} = 6\pi \times 0.529 \, \mathring{A}$.
$\lambda \approx 6 \times 3.1416 \times 0.529 \, \mathring{A} \approx 9.97 \, \mathring{A}$.
Since $1 \, \mathring{A} = 10^{-8} \, cm$,we have $\lambda = 9.96 \times 10^{-8} \, cm$.

(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Mass of one $H_2$ molecule $(m)$ = $\frac{2 \, g/mol}{6.023 \times 10^{23} \, mol^{-1}} \approx 3.32 \times 10^{-24} \, g$.
Given velocity $(v)$ = $5 \times 10^4 \, cm \cdot sec^{-1}$.
Planck's constant $(h)$ = $6.626 \times 10^{-27} \, erg \cdot sec$.
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-27} \, erg \cdot sec}{(3.32 \times 10^{-24} \, g) \times (5 \times 10^4 \, cm \cdot sec^{-1})}$
$\lambda = \frac{6.626 \times 10^{-27}}{16.6 \times 10^{-20}} \, cm \approx 3.99 \times 10^{-8} \, cm$.
Since $1 \, \mathring{A} = 10^{-8} \, cm$,the wavelength is approximately $4 \, \mathring{A}$.

(C) According to the de Broglie equation,$\lambda = \frac{h}{mv}$.
Given: $h = 6.625 \times 10^{-34} \ J \ s$,$m = 200 \ g = 0.2 \ kg$,and $v = 5 \ m/h = \frac{5}{3600} \ m/s$.
Substituting the values:
$\lambda = \frac{6.625 \times 10^{-34}}{0.2 \times (5 / 3600)} = \frac{6.625 \times 10^{-34}}{1 / 3600} = 6.625 \times 10^{-34} \times 3600 \approx 2.385 \times 10^{-30} \ m$.
Rounding to the nearest order of magnitude,the answer is $10^{-30} \ m$.

(C) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Given: $h = 6.626 \times 10^{-34} \ J \cdot s$,$m = 0.5 \ kg$,and $v = 100 \ m/s$.
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34}}{0.5 \times 100} = \frac{6.626 \times 10^{-34}}{50} = 1.325 \times 10^{-35} \ m$.
Thus,the correct option is $C$.

(C) The dual nature of matter (particles) was proposed by Louis de Broglie in $1924$.
He suggested that all matter,like radiation,exhibits both wave-like and particle-like properties.
The relationship between wavelength $(\lambda)$ and momentum $(p = mv)$ is given by the de Broglie equation:
$\lambda = \frac{h}{mv}$

(B) The speed of light $c = 3.00 \times 10^8 \ m \ s^{-1}$.
One percent of the speed of light is $v = (\frac{1}{100}) \times (3.00 \times 10^8 \ m \ s^{-1}) = 3.00 \times 10^6 \ m \ s^{-1}$.
Momentum of the electron $(p) = m \times v$.
$p = (9.11 \times 10^{-31} \ kg) \times (3.00 \times 10^6 \ m \ s^{-1}) = 2.733 \times 10^{-24} \ kg \ m \ s^{-1}$.
The de-Broglie wavelength $(\lambda)$ is given by $\lambda = \frac{h}{p}$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{2.733 \times 10^{-24} \ kg \ m \ s^{-1}} = 2.424 \times 10^{-10} \ m$.

(A) According to the de-Broglie hypothesis,the wavelength $(\lambda)$ associated with a particle of momentum $(p)$ is given by the equation: $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
Therefore,the correct relationship is $\lambda = \frac{h}{p}$.

(D) The de Broglie equation is given by $\lambda = \frac{h}{mv}$.
This equation relates the wavelength $\lambda$ to the momentum $p = mv$ of a particle.
It is a fundamental principle of quantum mechanics that applies to all material objects in motion,whether they are microscopic particles like electrons or macroscopic objects.

(A) According to Bohr's postulate, the circumference of an orbit is an integral multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.
For the $n^{th}$ orbit, the radius is given by $r_n = n^2 a_0$.
For the third orbit $(n = 3)$, the radius is $r_3 = 3^2 a_0 = 9 a_0$.
Substituting these values into the equation: $2 \pi (9 a_0) = 3 \lambda$.
$18 \pi a_0 = 3 \lambda$.
$\lambda = 6 \pi a_0$.

(A) According to the de Broglie hypothesis,a particle of mass $m$ moving with velocity $v$ has an associated wavelength $\lambda = \frac{h}{mv}$.
For an electron moving in a circular orbit of radius $r$,the circumference is $2\pi r$.
Bohr's quantization condition states that the angular momentum is quantized as $mvr = \frac{nh}{2\pi}$,which rearranges to $2\pi r = \frac{nh}{mv}$.
Substituting $\lambda = \frac{h}{mv}$ into the equation,we get $2\pi r = n\lambda$.
Thus,the circumference of the orbit must be an integral multiple of the de Broglie wavelength for a stable orbit.

(A) The kinetic energy $(K.E.)$ is given by $K.E. = \frac{p^2}{2m}$,where $p$ is momentum and $m$ is the mass of the electron.
From this,$p = \sqrt{2mE}$.
The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Given: $E = 2.5 \, eV = 2.5 \times 1.6 \times 10^{-12} \, erg$,$m = 9.1 \times 10^{-28} \, g$,and $h = 6.626 \times 10^{-27} \, erg \cdot s$.
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-27}}{\sqrt{2 \times 9.1 \times 10^{-28} \times 2.5 \times 1.6 \times 10^{-12}}}$
$\lambda = \frac{6.626 \times 10^{-27}}{\sqrt{72.8 \times 10^{-40}}}$
$\lambda = \frac{6.626 \times 10^{-27}}{8.53 \times 10^{-20}} \approx 7.7 \times 10^{-8} \, cm$.

(B) The de Broglie wavelength $(\lambda)$ is related to kinetic energy $(K)$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Since $\lambda \propto \frac{1}{\sqrt{K}}$,if the kinetic energy $K$ becomes $4K$,the new wavelength $\lambda'$ will be: $\lambda' = \frac{h}{\sqrt{2m(4K)}} = \frac{1}{2} \times \frac{h}{\sqrt{2mK}} = \frac{1}{2} \lambda$.
Therefore,the wavelength becomes half of its original value.

(A) According to the de Broglie equation,$p = \frac{h}{\lambda}$.
Given: $\lambda = 1 \, \mathring{A} = 1 \times 10^{-8} \, \text{cm}$.
$h = 6.6252 \times 10^{-27} \, \text{erg} \cdot \text{s}$.
Substituting the values: $p = \frac{6.6252 \times 10^{-27}}{1 \times 10^{-8}} \, \text{g} \cdot \text{cm/s}$.
$p = 6.6252 \times 10^{-19} \, \text{g} \cdot \text{cm/s}$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given: $m = 25 \, g$,$v = 6.6 \times 10^4 \, cm/sec$,$h = 6.626 \times 10^{-27} \, erg \cdot sec$ (in $CGS$ units).
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-27} \, erg \cdot sec}{25 \, g \times 6.6 \times 10^4 \, cm/sec}$
$\lambda \approx \frac{6.6 \times 10^{-27}}{25 \times 6.6 \times 10^4} \, cm$
$\lambda \approx \frac{10^{-27}}{25 \times 10^4} \, cm$
$\lambda \approx 0.04 \times 10^{-31} \, cm = 4 \times 10^{-33} \, cm$.

(D) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Substituting the given values:
$\lambda = \frac{6.63 \times 10^{-34} \, J s}{(1.67 \times 10^{-27} \, kg) \times (1.0 \times 10^3 \, m s^{-1})}$
$\lambda = \frac{6.63}{1.67} \times 10^{-34 + 27 - 3} \, m$
$\lambda \approx 3.97 \times 10^{-10} \, m$
To convert meters to nanometers $(1 \, nm = 10^{-9} \, m)$:
$\lambda = 3.97 \times 10^{-10} \, m = 0.397 \times 10^{-9} \, m \approx 0.40 \, nm$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $m = 60 \ g = 60 \times 10^{-3} \ kg = 0.06 \ kg$.
Velocity $v = 10 \ m/s$.
Planck's constant $h = 6.626 \times 10^{-34} \ J \cdot s$.
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{0.06 \times 10} = \frac{6.626 \times 10^{-34}}{0.6} \approx 1.1 \times 10^{-33} \ m$.
Thus,the value is approximately $10^{-33} \ m$.

(B) Given: Mass $m = 1 \, mg = 1 \times 10^{-3} \, g$,Velocity $v = 4.5 \times 10^5 \, cm/s$,Planck's constant $h = 6.652 \times 10^{-27} \, erg \cdot s$.
According to the de Broglie equation: $\lambda = \frac{h}{mv}$.
Substituting the values: $\lambda = \frac{6.652 \times 10^{-27} \, erg \cdot s}{(1 \times 10^{-3} \, g) \times (4.5 \times 10^5 \, cm/s)}$.
$\lambda = \frac{6.652 \times 10^{-27}}{4.5 \times 10^2} \, cm$.
$\lambda = 1.4782 \times 10^{-29} \, cm$ (approximately $1.4722 \times 10^{-29} \, cm$ based on provided options).

(B) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Given: $h = 6.626 \times 10^{-34} \, J \cdot s$,$m = 9.11 \times 10^{-31} \, kg$,and $v = 10^8 \, cm/s = 10^6 \, m/s$.
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s}{(9.11 \times 10^{-31} \, kg) \times (10^6 \, m/s)}$
$\lambda \approx 0.727 \times 10^{-9} \, m = 7.27 \times 10^{-10} \, m$.
Since $1 \, \mathop {\rm{A}}\limits^{\rm{o}} = 10^{-10} \, m$,the wavelength is approximately $7.27 \, \mathop {\rm{A}}\limits^{\rm{o}}$,which is closest to $7.25 \, \mathop {\rm{A}}\limits^{\rm{o}}$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \, J \cdot s$
$m = 9.1 \times 10^{-31} \, kg$
$v = 1.2 \times 10^5 \, ms^{-1}$
Substituting these values:
$\lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (1.2 \times 10^5)}$
$\lambda = \frac{6.626 \times 10^{-34}}{10.92 \times 10^{-26}}$
$\lambda \approx 0.6068 \times 10^{-8} \, m = 6.068 \times 10^{-9} \, m$.

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Since the velocity $v$ is assumed to be constant for comparison,$\lambda \propto \frac{1}{m}$.
The molar mass of $SO_2$ $(64 \ g/mol)$ is greater than the masses of $e^-$,$p$,and $CO_2$ $(44 \ g/mol)$.
Therefore,$SO_2$ has the highest mass,which results in the lowest de Broglie wavelength $\lambda$.

(A) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \ J \cdot s$
$m = 1.67 \times 10^{-27} \ kg$
$v = 10^3 \ m/s$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^3}$
$\lambda = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-24}}$
$\lambda \approx 3.967 \times 10^{-10} \ m$
To convert meters to nanometers $(nm)$,multiply by $10^9$:
$\lambda = 3.967 \times 10^{-10} \times 10^9 \ nm = 0.3967 \ nm$
Rounding off,we get $\lambda \approx 0.40 \ nm$.

(C) Given: Mass $m = 9.1 \times 10^{-31} \, kg$,Velocity $v = \frac{1}{10} \times c = \frac{1}{10} \times 3 \times 10^8 \, m/s = 3 \times 10^7 \, m/s$,Planck's constant $h = 6.626 \times 10^{-34} \, J \cdot s$.
Using the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$.
$\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s}{(9.1 \times 10^{-31} \, kg) \times (3 \times 10^7 \, m/s)}$.
$\lambda = \frac{6.626 \times 10^{-34}}{27.3 \times 10^{-24}} \, m$.
$\lambda \approx 0.2427 \times 10^{-10} \, m = 2.427 \times 10^{-11} \, m$.

(C) According to the de Broglie equation,$\lambda = \frac{h}{mv}$.
For methane $(CH_4)$,$\lambda_{CH_4} = \frac{h}{m_{CH_4} \times v_{CH_4}}$.
For oxygen $(O_2)$,$\lambda_{O_2} = \frac{h}{m_{O_2} \times v_{O_2}}$.
Since the wavelengths are equal,$\lambda_{CH_4} = \lambda_{O_2}$.
Therefore,$m_{CH_4} \times v_{CH_4} = m_{O_2} \times v_{O_2}$.
$\frac{v_{CH_4}}{v_{O_2}} = \frac{m_{O_2}}{m_{CH_4}} = \frac{32}{16} = 2$.
Thus,the ratio of velocities is $2 : 1$.

(A) According to Bohr's postulate,the angular momentum of an electron in a stationary orbit is quantized as $mvr = \frac{nh}{2\pi}$.
From de Broglie's hypothesis,the wavelength of an electron is $\lambda = \frac{h}{mv}$.
Substituting $mv = \frac{h}{\lambda}$ into the Bohr's postulate: $\frac{h}{\lambda} \times r = \frac{nh}{2\pi}$.
This simplifies to $2\pi r = n\lambda$,where $2\pi r$ is the circumference of the orbit.
Therefore,the number of waves formed by an electron in an orbit is equal to the principal quantum number $n$.
For the $n = 3$ orbit,the number of waves is $3$.

(B) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.
Given: $h = 6.626 \times 10^{-34} \ J \cdot s = 6.626 \times 10^{-27} \ erg \cdot s$,$m = 25 \ g$,$v = 6.6 \times 10^4 \ cm/sec$.
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-27} \ erg \cdot s}{25 \ g \times 6.6 \times 10^4 \ cm/sec}$
$\lambda = \frac{10^{-27}}{25 \times 10^4} \ cm$
$\lambda = \frac{1}{25} \times 10^{-31} \ cm$
$\lambda = 0.04 \times 10^{-31} \ cm = 4 \times 10^{-33} \ cm$.

(D) According to the De Broglie relation,the momentum $p = \frac{h}{\lambda}$.
Since $p = m \times c$,we have $m = \frac{h}{\lambda \times c}$.
Given: $h = 6.6252 \times 10^{-34} \, kg \cdot m^2/s$,$\lambda = 5894 \, \mathring{A} = 5894 \times 10^{-10} \, m$,and $c = 3 \times 10^8 \, m/s$.
Substituting the values:
$m = \frac{6.6252 \times 10^{-34}}{5894 \times 10^{-10} \times 3 \times 10^8}$
$m = \frac{6.6252 \times 10^{-34}}{17682 \times 10^{-2}}$
$m = 0.00037468 \times 10^{-32} \, kg$
$m = 3.746 \times 10^{-36} \, kg$.

(C) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 9.1 \times 10^{-31} \ kg$,$E = 2.8 \times 10^{-23} \ J$.
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.8 \times 10^{-23}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{50.96 \times 10^{-54}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{7.138 \times 10^{-27}}$
$\lambda \approx 9.28 \times 10^{-8} \ m$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For particle $X$: $\lambda_X = \frac{h}{m_X v_X} = 1 \ \mathring{A}$.
For particle $Y$: $m_Y = 0.25 m_X$ and $v_Y = 0.75 v_X$.
Therefore,$\lambda_Y = \frac{h}{m_Y v_Y} = \frac{h}{(0.25 m_X)(0.75 v_X)} = \frac{h}{0.1875 m_X v_X}$.
Substituting $\lambda_X = \frac{h}{m_X v_X} = 1$,we get $\lambda_Y = \frac{1}{0.1875} = \frac{1}{3/16} = \frac{16}{3} \approx 5.33 \ \mathring{A}$.

(D) According to the de Broglie relation,the wavelength $\lambda$ associated with a particle of momentum $p$ is given by the equation: $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
This equation shows that the wavelength $\lambda$ is inversely proportional to the momentum $p$ of the particle.
Therefore,the correct option is $D$.

(B) According to the de Broglie wavelength formula,$\lambda = \frac{h}{mv}$.
Since the velocity $(v)$ and Planck's constant $(h)$ are constant,$\lambda \propto \frac{1}{m}$.
Therefore,$\frac{\lambda_{He}}{\lambda_{H_2}} = \frac{m_{H_2}}{m_{He}}$.
The molar mass of $H_2$ is $2 \ g/mol$ and the molar mass of $He$ is $4 \ g/mol$.
Substituting the values: $\frac{\lambda_{He}}{\lambda_{H_2}} = \frac{2}{4} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(A) According to the de Broglie hypothesis,the wavelength $(\lambda)$ associated with a particle of momentum $(P)$ is given by the equation: $\lambda = \frac{h}{P}$,where $h$ is Planck's constant. This relationship is known as the de Broglie relation.