Matter Waves and de Broglie Wavelength Questions in English

(B) In $1924$,the French physicist Louis de-Broglie proposed the hypothesis that matter,like radiation,exhibits a dual nature (wave-particle duality). He suggested that every moving particle of matter is associated with a wave,which is known as a matter wave or de-Broglie wave. Therefore,the correct option is $B$.

(C) According to the de-Broglie hypothesis,every moving particle of matter is associated with a wave,known as a matter wave or de-Broglie wave.
The wavelength $\lambda$ of this wave is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.
Since the wavelength depends on the velocity $v$,the wave is associated with matter only when it is in motion (i.e.,$v \neq 0$).
Therefore,a wave is associated with matter when it is in motion with any velocity.

(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ associated with a particle of momentum $p$ is given by the relation: $\lambda = \frac{h}{p}$.
Since the momentum $p$ of a particle of mass $m$ moving with velocity $v$ is defined as $p = mv$,we substitute this into the equation.
Therefore,the de-Broglie wavelength is $\lambda = \frac{h}{mv}$.

(B) According to the de Broglie hypothesis,the wavelength $\lambda$ associated with a particle of momentum $p$ is given by $\lambda = \frac{h}{p}$.
Since the kinetic energy $E$ is related to momentum by $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
Here,$h$ is Planck's constant and $m$ is the mass of the electron,both of which are constants.
Therefore,$\lambda \propto \frac{1}{\sqrt{E}}$.
As the kinetic energy $E$ increases,the wavelength $\lambda$ must decrease.

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Given that the de-Broglie wavelengths for a proton $(p)$ and an $\alpha$-particle $(\alpha)$ are equal, we have $\lambda_p = \lambda_{\alpha}$.
Therefore, $\frac{h}{m_p v_p} = \frac{h}{m_{\alpha} v_{\alpha}}$, which simplifies to $m_p v_p = m_{\alpha} v_{\alpha}$.
The ratio of their velocities is $\frac{v_p}{v_{\alpha}} = \frac{m_{\alpha}}{m_p}$.
We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton, so $m_{\alpha} = 4 m_p$.
Substituting this, we get $\frac{v_p}{v_{\alpha}} = \frac{4 m_p}{m_p} = \frac{4}{1}$.
Thus, the ratio of their velocities is $4 : 1$.

(A) The kinetic energy $E$ of an electron of mass $m$ moving with velocity $v$ is given by $E = \frac{1}{2}mv^2$.
From this,we can write $v^2 = \frac{2E}{m}$,which implies $v = \sqrt{\frac{2E}{m}}$.
The momentum $p$ of the electron is $p = mv = m \sqrt{\frac{2E}{m}} = \sqrt{2mE}$.
According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.
Substituting the value of $p$,we get $\lambda = \frac{h}{\sqrt{2mE}}$.

(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
Since both the electron and the proton are accelerated through the same potential difference $V$ and both have the same magnitude of charge $q = e$,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.
For the electron: $\lambda_e = \lambda = \frac{k}{\sqrt{m}}$,where $k$ is a constant.
For the proton: $\lambda_p = \frac{k}{\sqrt{M}}$.
Dividing the two equations: $\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m}}{\sqrt{M}} = \sqrt{\frac{m}{M}}$.
Therefore,$\lambda_p = \lambda \sqrt{\frac{m}{M}}$.

(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.
For an $\alpha$-particle,the charge $q = 2e = 2 \times 1.6 \times 10^{-19} \ C$ and mass $m = 4m_p = 4 \times 1.67 \times 10^{-27} \ kg$.
Substituting these values:
$\lambda = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 4 \times 1.67 \times 10^{-27} \times 2 \times 1.6 \times 10^{-19} \times V}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{42.752 \times 10^{-46} \times V}} = \frac{6.63 \times 10^{-34}}{6.538 \times 10^{-23} \sqrt{V}} \ m$
$\lambda \approx \frac{1.014 \times 10^{-11}}{\sqrt{V}} \ m = \frac{0.1014 \times 10^{-10}}{\sqrt{V}} \ m = \frac{0.101}{\sqrt{V}} \ \mathring{A}$.

(B) The de-Broglie hypothesis proposed that matter,such as electrons,exhibits wave-like properties in addition to particle-like properties. This is known as wave-particle duality. Therefore,the hypothesis specifically introduced the concept of matter waves associated with electrons.

(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
Given values are:
$h = 6.6 \times 10^{-34} \ J \cdot s$
$m = 9 \times 10^{-31} \ kg$
$E = 80 \ eV = 80 \times 1.6 \times 10^{-19} \ J = 128 \times 10^{-19} \ J = 1.28 \times 10^{-17} \ J$.
Substituting these values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.28 \times 10^{-17}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{23.04 \times 10^{-48}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{4.8 \times 10^{-24}}$
$\lambda = 1.375 \times 10^{-10} \ m \approx 1.375 \ \mathring{A} \approx 1.4 \ \mathring{A}$.
Thus,the correct option is $D$.

(C) The de-Broglie wavelength $\lambda$ 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.
Given that the velocity $v$ is the same for all particles,the relationship becomes $\lambda \propto \frac{1}{m}$.
This means that the particle with the smallest mass will have the maximum de-Broglie wavelength.
Comparing the masses: $m_{\beta} < m_{proton} \approx m_{neutron} < m_{\alpha}$.
Since the $\beta$-particle (an electron) has the smallest mass among the given options,it will have the maximum de-Broglie wavelength.

(B) According to the de-Broglie hypothesis,the momentum $p$ of a particle is related to its wavelength $\lambda$ by the equation $p = \frac{h}{\lambda}$,where $h$ is Planck's constant.
Since both the electron and the photon have the same wavelength $\lambda$,and $h$ is a universal constant,their momenta must be equal.
Therefore,they have the same momentum.

(C) The de-Broglie wavelength $\lambda$ of a particle is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the particle.
From this equation,it is clear that the wavelength $\lambda$ is inversely proportional to the momentum $p$ of the particle.
Therefore,$\lambda \propto \frac{1}{p}$.

(A) The de Broglie wavelength of an electron is given by the formula $\lambda = \frac{h}{m_e v}$.
Rearranging the formula to solve for velocity $v$,we get $v = \frac{h}{m_e \lambda}$.
Substituting the known values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m_e = 9.11 \times 10^{-31} \ kg$,and $\lambda = 10^{-10} \ m$.
$v = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 10^{-10}} \ m/s$.
$v = \frac{6.63}{9.11} \times 10^{-34 + 31 + 10} \ m/s$.
$v \approx 0.7277 \times 10^7 \ m/s = 7.277 \times 10^6 \ m/s$.
Rounding to the nearest provided option,the speed is $7.25 \times 10^6 \ m/s$.

(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Here,$h = 6.63 \times 10^{-34} \ J \cdot s$ (Planck's constant).
The mass of a hydrogen molecule $(H_2)$ is $m = 2 \times 1.67 \times 10^{-27} \ kg = 3.34 \times 10^{-27} \ kg$.
The velocity is $v = 3 \ km/s = 3 \times 10^3 \ m/s$.
Substituting these values into the formula:
$\lambda = \frac{6.63 \times 10^{-34}}{3.34 \times 10^{-27} \times 3 \times 10^3} \ m$.
$\lambda = \frac{6.63 \times 10^{-34}}{10.02 \times 10^{-24}} \ m \approx 0.66 \times 10^{-10} \ m$.
Since $1 \ \mathring{A} = 10^{-10} \ m$,we get $\lambda = 0.66 \ \mathring{A}$.

(A) The de-Broglie wavelength of a neutron is given by $\lambda = \frac{h}{\sqrt{2mkT}}$,where $m$ is the mass of the neutron,$k$ is the Boltzmann constant,and $T$ is the absolute temperature.
Since $h$,$m$,and $k$ are constants,we have $\lambda \propto \frac{1}{\sqrt{T}}$.
Therefore,$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$.
Given $T_1 = 27^oC = 27 + 273 = 300 \ K$ and $T_2 = 927^oC = 927 + 273 = 1200 \ K$.
Substituting the values: $\frac{\lambda}{\lambda_2} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Thus,$\lambda_2 = \frac{\lambda}{2}$.

(D) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E$ is the kinetic energy.
Since the de-Broglie wavelength $\lambda$ is the same for both the electron and the proton,we have $\lambda = \text{constant}$.
From the formula,we can see that $E = \frac{h^2}{2m\lambda^2}$,which implies $E \propto \frac{1}{m}$.
Since the mass of an electron $(m_e)$ is much smaller than the mass of a proton $(m_p)$,i.e.,$m_e < m_p$,it follows that the kinetic energy of the electron $(E_e)$ must be greater than the kinetic energy of the proton $(E_p)$.
Therefore,$E_e > E_p$.

(B) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ moving with momentum $p$ is given by $\lambda = \frac{h}{p}$.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
While the de-Broglie wavelength exists for all matter,for a macroscopic object like a cricket ball,the wavelength is extremely small (on the order of $10^{-34} \ m$),making it physically undetectable and practically irrelevant,but the mathematical expression remains valid.

(A) For a photon,the energy $E$ is given by $E = \frac{hc}{\lambda_{Ph}}$,so $\lambda_{Ph} = \frac{hc}{E}$.
For a non-relativistic electron,the energy $E$ is given by $E = \frac{p^2}{2m}$,where $p$ is the momentum. The de Broglie wavelength is $\lambda_{el} = \frac{h}{p}$.
From the energy equation,$p = \sqrt{2mE}$,so $\lambda_{el} = \frac{h}{\sqrt{2mE}}$.
Comparing the two expressions:
$\lambda_{Ph} = \frac{hc}{E}$
$\lambda_{el} = \frac{h}{\sqrt{2mE}}$
Given the same energy $E$,the ratio is $\frac{\lambda_{Ph}}{\lambda_{el}} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{c\sqrt{2mE}}{E} = c \sqrt{\frac{2m}{E}}$.
Since $c$ is the speed of light $(3 \times 10^8 \ m/s)$ and $E$ is very small $(10^{-20} \ J)$,the value of $\lambda_{Ph}$ is significantly larger than $\lambda_{el}$.
Therefore,$\lambda_{Ph} > \lambda_{el}$.

(B) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $K$ of an electron by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Rearranging for $K$, we get: $K = \frac{h^2}{2m\lambda^2}$.
Given $\lambda = 0.3 \, nm = 0.3 \times 10^{-9} \, m$, $h = 6.626 \times 10^{-34} \, J \cdot s$, and $m = 9.1 \times 10^{-31} \, kg$.
Substituting the values: $K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (0.3 \times 10^{-9})^2}$.
$K = \frac{43.9 \times 10^{-68}}{18.2 \times 10^{-31} \times 0.09 \times 10^{-18}} = \frac{43.9 \times 10^{-68}}{1.638 \times 10^{-48}} \approx 26.8 \times 10^{-20} \, J$.
To convert Joules to $eV$, divide by $1.6 \times 10^{-19} \, J/eV$: $K \approx \frac{26.8 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 16.75 \, eV \approx 16.8 \, eV$.

(A) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ by the equation: $\lambda = \frac{h}{p}$.
Rearranging the formula to solve for momentum $p$,we get: $p = \frac{h}{\lambda}$.
Given values are: $h = 6.63 \times 10^{-34} \ J \cdot s$ and $\lambda = 2 \mu m = 2 \times 10^{-6} \ m$.
Substituting these values into the equation:
$p = \frac{6.63 \times 10^{-34}}{2 \times 10^{-6}} \ kg \cdot m/s$.
$p = 3.315 \times 10^{-28} \ kg \cdot m/s$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 1 \ kg$,$v = 2000 \ m/s$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34}}{1 \times 2000} = \frac{6.63}{2} \times 10^{-37} \ m = 3.315 \times 10^{-37} \ m$.
Since $1 \ \mathring{A} = 10^{-10} \ m$,we convert the wavelength to $\mathring{A}$:
$\lambda = 3.315 \times 10^{-37} \ m = 3.315 \times 10^{-27} \ \mathring{A}$.
Rounding to one decimal place,we get $3.3 \times 10^{-27} \ \mathring{A}$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.
Given: $h = 6.6 \times 10^{-34} \ J \cdot s$,$m = 9.1 \times 10^{-31} \ kg$,and $E = 5 \ eV = 5 \times 1.6 \times 10^{-19} \ J$.
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{-19}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{145.6 \times 10^{-50}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{12.066 \times 10^{-25}}$
$\lambda \approx 0.547 \times 10^{-9} \ m = 5.47 \times 10^{-10} \ m = 5.47 \ \mathring{A}$.

(C) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mQV}}$.
Substituting the values $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 9.11 \times 10^{-31} \ kg$,$Q = 1.60 \times 10^{-19} \ C$,and $V = 100 \ V$:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 100}}$
$\lambda \approx \frac{12.27}{\sqrt{V}} \ \mathring A$
For $V = 100 \ V$,$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \ \mathring A \approx 1.23 \ \mathring A$.

(C) The de-Broglie wavelength $\lambda$ of a particle is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the particle.
From this relation,it is clear that $\lambda \propto \frac{1}{p}$.
Therefore,the de-Broglie wavelength is inversely proportional to the momentum of the particle.
Thus,option $C$ is correct.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
From this expression,it is clear that $\lambda \propto \frac{1}{\sqrt{E}}$.
If the kinetic energy $E$ becomes $E' = 2E$,the new wavelength $\lambda'$ is given by $\lambda' = \frac{h}{\sqrt{2m(2E)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mE}}$.
Therefore,$\lambda' = \frac{1}{\sqrt{2}} \lambda$.
Thus,the wavelength changes by a factor of $1/\sqrt{2}$.

(D) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ of a particle by the equation $\lambda = \frac{h}{p}$.
We know that the kinetic energy $E$ of a body of mass $m$ is given by $E = \frac{p^2}{2m}$.
Rearranging this for momentum,we get $p^2 = 2mE$,which implies $p = \sqrt{2mE}$.
Substituting the expression for $p$ into the de-Broglie wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
Therefore,the correct option is $D$.

(D) The wavelength of a matter wave (de Broglie wavelength) is given by the formula:
$\lambda = \frac{h}{mv} = \frac{h}{p}$ $....(i)$
where $h$ is Planck's constant,$m$ is the mass of the particle,$v$ is its velocity,and $p$ is its momentum.
From equation $(i)$,it is clear that the wavelength $\lambda$ depends on the mass,velocity,and momentum of the particle.
It does not depend on the charge of the particle.
Therefore,the correct option is $D$.

(C) According to the law of conservation of linear momentum,the initial momentum of the system is zero.
$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2$
$\Rightarrow m_1 \vec{v}_1 = -m_2 \vec{v}_2$
Taking the magnitude,we get $m_1 v_1 = m_2 v_2$,which implies that the magnitudes of the momenta of the two particles are equal $(p_1 = p_2 = p)$.
The de-Broglie wavelength is given by $\lambda = h / p$.
Since both particles have the same magnitude of momentum $p$,their de-Broglie wavelengths will be $\lambda_1 = h / p$ and $\lambda_2 = h / p$.
Therefore,the ratio $\lambda_1 / \lambda_2 = (h / p) / (h / p) = 1$.

(B) For a photon,the energy is given by $E = \frac{hc}{\lambda_{\text{photon}}}$,which implies $\lambda_{\text{photon}} = \frac{hc}{E}$.
For an electron,the de Broglie wavelength is given by $\lambda_{\text{electron}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $m$ is the mass of the electron and $E$ is its kinetic energy.
Taking the ratio of the two wavelengths:
$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{hc}{E} \cdot \frac{\sqrt{2mE}}{h} = c \sqrt{\frac{2m}{E}}$.
Since $c$ and $m$ are constants,we have $\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} \propto \frac{1}{\sqrt{E}}$.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{rms}}$.
For a gas molecule,the root mean square velocity is $v_{rms} = \sqrt{\frac{3kT}{m}}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{3mkT}}$.
Given $T_H = 27^oC = 300 K$ and $T_{He} = 127^oC = 400 K$.
The masses are $m_H = 2$ units and $m_{He} = 4$ units.
The ratio is $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{m_{He} T_{He}}{m_H T_H}} = \sqrt{\frac{4 \times 400}{2 \times 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{8}{3}}$.

(C) The total energy of the particle is $E = 2 E_0$.
For the region $0 \le x \le 1$,the potential energy is $U_1 = E_0$. The kinetic energy is $K_1 = E - U_1 = 2 E_0 - E_0 = E_0$.
The de-Broglie wavelength is $\lambda_1 = \frac{h}{\sqrt{2m K_1}} = \frac{h}{\sqrt{2m E_0}}$.
For the region $x > 1$,the potential energy is $U_2 = 0$. The kinetic energy is $K_2 = E - U_2 = 2 E_0 - 0 = 2 E_0$.
The de-Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2m K_2}} = \frac{h}{\sqrt{2m(2 E_0)}} = \frac{h}{\sqrt{4m E_0}}$.
Taking the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2m E_0}}{h / \sqrt{4m E_0}} = \sqrt{\frac{4m E_0}{2m E_0}} = \sqrt{2}$.

(C) The de-Broglie wavelength $\lambda$ of an electron with energy $E$ is given by the relation:
$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m}} \cdot E^{-1/2}$
Taking the logarithm on both sides:
$\log \lambda = \log \left( \frac{h}{\sqrt{2m}} \cdot E^{-1/2} \right)$
Using the properties of logarithms:
$\log \lambda = \log \left( \frac{h}{\sqrt{2m}} \right) + \log (E^{-1/2})$
$\log \lambda = -\frac{1}{2} \log E + \log \left( \frac{h}{\sqrt{2m}} \right)$
This equation is in the form of a straight line $y = mx + c$,where:
$y = \log \lambda$
$x = \log E$
$m = -1/2$ (slope)
$c = \log \left( \frac{h}{\sqrt{2m}} \right)$ (positive intercept on the $y$-axis)
Since the slope is negative,the graph is a straight line with a negative slope,which corresponds to the graph shown in option $C$.

(B) For an electron confined between two rigid walls of length $L$,the particle acts like a standing wave. For the lowest energy state (ground state),the length $L$ corresponds to half a wavelength: $L = \frac{\lambda}{2}$,so $\lambda = 2L$.
Given $L = 1.0 \times 10^{-9} \ m$,we have $\lambda = 2 \times 10^{-9} \ m$.
Using the de Broglie relation $p = \frac{h}{\lambda}$,the kinetic energy $E$ is given by $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 9.1 \times 10^{-31} \ kg$,and $\lambda = 2 \times 10^{-9} \ m$:
$E = \frac{(6.63 \times 10^{-34})^2}{2 \times (9.1 \times 10^{-31}) \times (2 \times 10^{-9})^2}$
$E = \frac{43.96 \times 10^{-68}}{18.2 \times 10^{-31} \times 4 \times 10^{-18}}$
$E = \frac{43.96 \times 10^{-68}}{72.8 \times 10^{-49}} \approx 0.6038 \times 10^{-19} \ J = 6.038 \times 10^{-20} \ J$.
Thus,the closest value is $6.0 \times 10^{-20} \ J$.

(B) The wavelength of the electron wave must be $\lambda = 10 \ pm = 10 \times 10^{-12} \ m$.
Using the de Broglie wavelength formula: $\lambda = \frac{h}{\sqrt{2mE}}$, we can solve for energy $E$:
$E = \frac{h^2}{2m\lambda^2}$
Substituting the values $h = 6.63 \times 10^{-34} \ J \cdot s$, $m = 9.1 \times 10^{-31} \ kg$, and $\lambda = 10^{-11} \ m$:
$E = \frac{(6.63 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (10^{-11})^2} \ J$
$E \approx 2.41 \times 10^{-15} \ J$
To convert this energy into $eV$, divide by $1.6 \times 10^{-19} \ J/eV$:
$E = \frac{2.41 \times 10^{-15}}{1.6 \times 10^{-19}} \ eV \approx 15062 \ eV \approx 15 \ KeV$.

(C) Let the two particles have charges $q_1$ and $q_2$. Given $q_1 = q_2 = q$. Both are accelerated through the same potential difference $V$.
The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through potential $V$ is $K = qV$.
Since $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK} = \sqrt{2mqV}$.
The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.
Since $h$,$q$,and $V$ are constant for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,the ratio of their wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}}$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{3mk_BT}}$.
Here,$m = 9.11 \times 10^{-31} \ kg$,$k_B = 1.38 \times 10^{-23} \ J/K$,and $T = 27 + 273 = 300 \ K$.
Substituting these values:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times (9.11 \times 10^{-31}) \times (1.38 \times 10^{-23}) \times 300}}$
$\lambda \approx 6.2 \times 10^{-9} \ m$.
The mean distance between two electrons is $r = 2 \times 10^{-10} \ m$.
The ratio is $\frac{r}{\lambda} = \frac{2 \times 10^{-10}}{6.2 \times 10^{-9}} \approx 0.03$.

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,which implies $\lambda \propto \frac{1}{p}$.
Given that the wavelength changes by $0.25\%$,the new wavelength $\lambda' = \lambda + \frac{0.25}{100}\lambda = 1.0025\lambda$.
Since the wavelength increases,the momentum must decrease. Let the new momentum be $p' = p - p_0$.
Using the relation $\lambda' = \frac{h}{p'}$,we have $1.0025\lambda = \frac{h}{p - p_0}$.
Dividing this by $\lambda = \frac{h}{p}$,we get $1.0025 = \frac{p}{p - p_0}$.
$1.0025(p - p_0) = p$.
$1.0025p - 1.0025p_0 = p$.
$0.0025p = 1.0025p_0$.
$p = \frac{1.0025}{0.0025}p_0 = 401\ p_0$.

(A) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{3mk_BT}}$
Given values are:
Mass of electron,$m = 9.11 \times 10^{-31} \ kg$
Boltzmann constant,$k_B = 1.38 \times 10^{-23} \ J/K$
Temperature,$T = 27 + 273 = 300 \ K$
Planck's constant,$h = 6.63 \times 10^{-34} \ J \cdot s$
Substituting these values into the formula:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times (9.11 \times 10^{-31}) \times (1.38 \times 10^{-23}) \times 300}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{11293.62 \times 10^{-54}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{1.0627 \times 10^{-25}}$
$\lambda \approx 6.24 \times 10^{-9} \ m$
Thus,the correct option is $A$.

(A) 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 $(p)$: $m_p = m$, $q_p = e$.
For an $\alpha$-particle $(\alpha)$: $m_{\alpha} = 4m$, $q_{\alpha} = 2e$.
The ratio of the wavelengths is:
$\frac{\lambda_p}{\lambda_{\alpha}} = \frac{h / \sqrt{2m_p q_p V}}{h / \sqrt{2m_{\alpha} q_{\alpha} V}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$.
Substituting the values:
$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m \times 2e}{m \times e}} = \sqrt{8} = 2\sqrt{2}$.
Thus, the ratio is $2\sqrt{2} : 1$.

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Since the velocity $v$ is the same for all particles,$\lambda \propto \frac{1}{m}$.
This means the particle with the smallest mass will have the maximum de Broglie wavelength.
The masses of the given particles are in the order: $m_{\alpha} > m_n \approx m_p > m_{\beta}$.
Since the $\beta$-particle (electron) has the smallest mass among the given options,it will have the maximum de Broglie wavelength.

(C) According to the de Broglie hypothesis,every moving particle,whether microscopic (like an electron) or macroscopic (like a ball),is associated with a wave known as a matter wave or de Broglie wave.
$1$. Statement $A$ is correct: de Broglie waves represent the probability of finding a particle and are not physical waves like sound or water waves.
$2$. Statement $B$ is correct: The de Broglie wavelength $\lambda$ is given by $\lambda = h/p$,where $h$ is Planck's constant and $p$ is momentum. Thus,$\lambda \propto 1/p$.
$3$. Statement $C$ is incorrect: The wave nature is associated with all particles,regardless of their size (atomic or macroscopic). However,for macroscopic objects,the wavelength is too small to be detected.
$4$. Statement $D$ is correct: Due to the extremely small value of Planck's constant $h$,the wavelength of macroscopic objects is negligible,making the wave nature unobservable in daily life.
Therefore,the incorrect statement is $C$.

(A) The kinetic energy of a neutron at temperature $T$ is given by $K.E. = \frac{3}{2}kT = \frac{p^2}{2m}$.
Thus,the momentum $p = \sqrt{3mkT}$.
The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}$.
This implies $\lambda \propto \frac{1}{\sqrt{T}}$.
Given $T_1 = 27^{\circ}C = 300 \ K$ and $T_2 = 927^{\circ}C = 1200 \ K$.
Therefore,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Hence,$\lambda_2 = \frac{\lambda}{2}$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.
Taking the derivative, we get $d\lambda = -\frac{h}{p^2} dp$.
Dividing by $\lambda = \frac{h}{p}$, we obtain $\frac{d\lambda}{\lambda} = -\frac{dp}{p}$.
Taking the magnitude, $\frac{|d\lambda|}{\lambda} = \frac{|dp|}{p}$.
Given $\frac{|d\lambda|}{\lambda} \times 100 = 0.50\ \%$, so $\frac{|d\lambda|}{\lambda} = \frac{0.50}{100} = \frac{1}{200}$.
Given the change in momentum $|dp| = P_m$, we have $\frac{P_m}{p} = \frac{1}{200}$.
Therefore, the initial momentum $p = 200\ P_m$.

(A) The de Broglie wavelength $\lambda$ is related to momentum $p$ as $\lambda = \frac{h}{p}$.
For a particle with mass $m$ and kinetic energy $E$,the momentum is $p = \sqrt{2mE}$,so $\lambda = \frac{h}{\sqrt{2mE}}$.
Rearranging for energy,we get $E = \frac{h^2}{2m\lambda^2}$.
For a photon,the energy is $E = \frac{hc}{\lambda}$.
Comparing the two,since the mass $m$ of the uranium nucleus is much larger than the mass of the electron,and the photon's energy expression is inversely proportional to $\lambda$ (rather than $\lambda^2$),the photon has the highest energy for a given wavelength $\lambda$.

(A) According to the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the particle.
For a particle moving at relativistic speeds $(v \approx c)$,the mass $m$ is given by the relativistic mass formula: $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$,where $m_0$ is the rest mass.
The momentum $p$ is defined as $p = m \cdot v$.
Substituting the expression for relativistic mass into the momentum formula,we get: $p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}$.
Finally,substituting this into the de Broglie wavelength equation,we obtain: $\lambda = \frac{h}{m_0 v / \sqrt{1 - \frac{v^2}{c^2}}}$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its velocity.
From this relation,we can see that $\lambda \propto \frac{1}{v}$.
This implies that the de Broglie wavelength is inversely proportional to the velocity of the electron.
Therefore,if the velocity $v$ of the electron increases,the de Broglie wavelength $\lambda$ will decrease.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Since $h$ and $m$ are constants,$\lambda \propto \frac{1}{v}$.
Let the initial velocity be $v_1 = v$ and the initial wavelength be $\lambda_1 = \lambda$.
When the velocity is doubled,the new velocity is $v_2 = 2v$.
The new wavelength is $\lambda_2 = \frac{h}{m(2v)} = \frac{1}{2} \left( \frac{h}{mv} \right) = \frac{\lambda}{2}$.
The change in wavelength is $\Delta \lambda = \lambda_1 - \lambda_2 = \lambda - \frac{\lambda}{2} = \frac{\lambda}{2}$.
Thus,the wavelength decreases by $\frac{\lambda}{2}$.