Matter Waves and de Broglie Wavelength Questions in English

(C) The de Broglie wavelength of an electron with kinetic energy $K$ is given by $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.
For an electron,$\lambda_e \approx \frac{12.27}{\sqrt{K}} \, \mathring{A}$,where $K$ is in $eV$.
Given $K = 1.5 \, eV$,we have $\lambda_e = \frac{12.27}{\sqrt{1.5}} \, \mathring{A}$.
The wavelength of a photon is given by $\lambda_p = \frac{hc}{E_p}$.
Since $\lambda_p = \lambda_e$,we have $\frac{hc}{E_p} = \frac{12.27}{\sqrt{1.5}} \, \mathring{A}$.
Using $hc \approx 12400 \, eV \cdot \mathring{A}$,we get $E_p = \frac{12400 \cdot \sqrt{1.5}}{12.27} \, eV$.
$E_p \approx \frac{12400 \cdot 1.2247}{12.27} \approx 1237.5 \, eV \approx 1.24 \, keV$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $K$ is the kinetic energy.
From this relation,we see that $\lambda \propto \frac{1}{\sqrt{K}}$.
Let the initial kinetic energy be $K_1 = K$ and the final kinetic energy be $K_2 = 16K$.
Then,the ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K}{16K}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 0.25$.
So,$\lambda_2 = 0.25 \lambda_1$.
The percentage change in wavelength is given by $\frac{\lambda_1 - \lambda_2}{\lambda_1} \times 100\%$.
Substituting the values,we get $\frac{\lambda_1 - 0.25 \lambda_1}{\lambda_1} \times 100\% = 0.75 \times 100\% = 75\%$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$, where $h$ is Planck's constant, $m$ is the mass of the particle, and $K$ is the kinetic energy.
From this relation, we see that $\lambda \propto \frac{1}{\sqrt{K}}$.
Let the initial kinetic energy be $K_1 = K$ and the final kinetic energy be $K_2 = 16K$.
The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K}{16K}} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 0.25$.
Thus, $\lambda_2 = 0.25\lambda_1$.
The percentage change in wavelength is given by $\frac{\lambda_1 - \lambda_2}{\lambda_1} \times 100\%$.
Substituting the values, we get $\frac{\lambda_1 - 0.25\lambda_1}{\lambda_1} \times 100\% = 0.75 \times 100\% = 75\%$.

(A) The electron enters the region between the plates with an initial kinetic energy of $200 \, eV$.
Since the plates are connected to a $100 \, V$ power supply, there is an electric field between them. Based on the diagram, the electron moves against the electric field (from positive to negative plate), which acts as a retarding potential.
Therefore, the final kinetic energy of the electron when it emerges from the second plate is $K_f = 200 \, eV - 100 \, eV = 100 \, eV$.
The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A}$, where $V$ is the kinetic energy in $eV$.
Substituting $V = 100 \, eV$:
$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \, \mathring{A} \approx 1.22 \, \mathring{A}$.

(D) The de Broglie wavelength $\lambda$ is related to the kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
From this,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E'}}$.
Given $\lambda = 1 \, nm$ and $\lambda' = 0.5 \, nm$,we have $\frac{0.5}{1} = \sqrt{\frac{E}{E'}}$.
Squaring both sides,we get $\frac{1}{4} = \frac{E}{E'}$,which means $E' = 4E$.
The energy that should be added is $\Delta E = E' - E = 4E - E = 3E$.
Therefore,the energy to be added is thrice the initial energy.

(C) Let the momenta of the two electrons be $\vec{p}_1 = \frac{h}{\lambda_1} \hat{i}$ and $\vec{p}_2 = \frac{h}{\lambda_2} \hat{j}$.
Since both are electrons,they have the same mass $m$.
The velocity of the centre of mass is $\vec{V}_{CM} = \frac{\vec{p}_1 + \vec{p}_2}{2m} = \frac{h}{2m\lambda_1} \hat{i} + \frac{h}{2m\lambda_2} \hat{j}$.
The velocity of the first electron relative to the centre of mass is $\vec{v}_{1,CM} = \vec{v}_1 - \vec{V}_{CM} = \frac{\vec{p}_1}{m} - \frac{\vec{p}_1 + \vec{p}_2}{2m} = \frac{\vec{p}_1 - \vec{p}_2}{2m} = \frac{h}{2m\lambda_1} \hat{i} - \frac{h}{2m\lambda_2} \hat{j}$.
The momentum of the electron in the $CM$ frame is $\vec{p}_{CM} = m \vec{v}_{1,CM} = \frac{h}{2\lambda_1} \hat{i} - \frac{h}{2\lambda_2} \hat{j}$.
The magnitude of this momentum is $p_{CM} = \sqrt{(\frac{h}{2\lambda_1})^2 + (-\frac{h}{2\lambda_2})^2} = \frac{h}{2} \sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}} = \frac{h}{2} \frac{\sqrt{\lambda_1^2 + \lambda_2^2}}{\lambda_1 \lambda_2}$.
The de Broglie wavelength in the $CM$ frame is $\lambda_{CM} = \frac{h}{p_{CM}} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$.

(C) According to the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,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 simplifies to $m_p v_p = m_{\alpha} v_{\alpha}$,or $\frac{v_p}{v_{\alpha}} = \frac{m_{\alpha}}{m_p}$.
Since the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton $(m_{\alpha} = 4m_p)$,we substitute this into the ratio:
$\frac{v_p}{v_{\alpha}} = \frac{4m_p}{m_p} = \frac{4}{1}$.
Thus,the ratio of the velocities of the proton and the $\alpha$-particle is $4:1$.

(C) The de-Broglie wavelength $\lambda$ for an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}}\,\mathring{A}$.
Given $V = 50\,V$.
Substituting the value of $V$ in the formula:
$\lambda = \frac{12.27}{\sqrt{50}}\,\mathring{A}$.
Since $\sqrt{50} \approx 7.071$.
$\lambda = \frac{12.27}{7.071} \approx 1.735\,\mathring{A}$.
Rounding to one decimal place,we get $\lambda \approx 1.7\,\mathring{A}$.
Therefore,the correct option is $C$.

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For a particle to exhibit wave-like properties,its wavelength $\lambda$ must be large enough to be detected experimentally.
Since $\lambda$ is inversely proportional to the mass $m$ of the particle,particles with very large masses have extremely small wavelengths that are impossible to measure with current technology.
Among the given options,a dust particle has the largest mass ($m$ is very high).
Therefore,the de Broglie wavelength for a dust particle is extremely small,making it the most difficult to experimentally verify.

(C) According to the condition given,the circumference of the orbit is an integer multiple of the de-Broglie wavelength:
$2 \pi r = n \lambda$
Since $\lambda = \frac{h}{p} = \frac{h}{mv}$,we have:
$2 \pi r = \frac{nh}{mv} \implies mvr = \frac{nh}{2 \pi}$
For a charged particle moving in a magnetic field $B$,the magnetic Lorentz force provides the centripetal force:
$qvB = \frac{mv^2}{r} \implies mv = qBr$
Substituting $mv = qBr$ into the quantization condition:
$(qBr)r = \frac{nh}{2 \pi} \implies qBr^2 = \frac{nh}{2 \pi}$
Solving for $r^2$:
$r^2 = \frac{nh}{2 \pi qB}$
Thus,$r = \sqrt{\frac{nh}{2 \pi qB}}$,which implies $r \propto n^{1/2}$.

(D) The kinetic energy $E$ acquired by a charged particle accelerated through a potential difference $V$ is given by $E = qV$.
For an electron,$E_e = eV$.
For a proton,$E_p = e(4V) = 4eV$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Thus,$\lambda_e = \frac{h}{\sqrt{2m_e eV}}$ and $\lambda_p = \frac{h}{\sqrt{2m_p (4eV)}} = \frac{h}{2\sqrt{2m_p eV}}$.
Taking the ratio $\frac{\lambda_e}{\lambda_p}$:
$\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2m_e eV}} \times \frac{2\sqrt{2m_p eV}}{h} = 2\sqrt{\frac{m_p}{m_e}}$.

(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.
For relativistic electrons, the energy $E$ is related to momentum $p$ by $E^2 = p^2c^2 + m_0^2c^4$. However, for high energy, $E \approx pc = \frac{hc}{\lambda}$.
Given $\lambda = 7.5 \times 10^{-12} \ m$.
$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{7.5 \times 10^{-12}} \ J$.
$E = \frac{19.89 \times 10^{-26}}{7.5 \times 10^{-12}} \ J = 2.652 \times 10^{-14} \ J$.
Converting to $eV$: $E = \frac{2.652 \times 10^{-14}}{1.6 \times 10^{-19}} \ eV \approx 1.65 \times 10^5 \ eV = 165 \ keV$.
Considering the non-relativistic approximation provided in the prompt's logic $(V \approx 25 \ kV)$, the question implies the use of the de Broglie relation $\lambda = \frac{h}{\sqrt{2mE}}$.
$E = \frac{h^2}{2m\lambda^2} = \frac{(6.63 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (7.5 \times 10^{-12})^2} \ J$.
$E \approx 4.3 \times 10^{-15} \ J \approx 26.8 \ keV$.
Thus, the closest option is $25 \ keV$.

(D) The de Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{mv}$.
The wavelength of a photon is given by $\lambda_p = \frac{c}{\nu}$,where $\nu$ is the frequency.
According to the problem,$\lambda_e = 10^{-3} \times \lambda_p$.
Substituting the expressions: $\frac{h}{mv} = 10^{-3} \times \frac{c}{\nu}$.
Rearranging for the speed of the electron $(v)$: $v = \frac{h \nu}{m c \times 10^{-3}}$.
Substituting the given values: $v = \frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^8 \times 10^{-3}}$.
$v = \frac{39.78 \times 10^{-20}}{27.3 \times 10^{-26}} = 1.457 \times 10^6 \, m/s \approx 1.45 \times 10^6 \, m/s$.

(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For particle $A$: $\lambda_A = \frac{h}{\sqrt{2mq(50)}}$.
For particle $B$: $\lambda_B = \frac{h}{\sqrt{2(4m)q(2500)}} = \frac{h}{\sqrt{8mq(2500)}} = \frac{h}{\sqrt{20000mq}}$.
The ratio $\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{2(4m)q(2500)}}{\sqrt{2mq(50)}} = \sqrt{\frac{8m \cdot q \cdot 2500}{2m \cdot q \cdot 50}} = \sqrt{\frac{20000}{100}} = \sqrt{200} = 10\sqrt{2}$.
Since $\sqrt{2} \approx 1.414$,the ratio is $10 \times 1.414 = 14.14$.

(D) Let the momenta of the two particles be $\vec{P}_1$ and $\vec{P}_2$. Given they move at right angles,we can represent them as $\vec{P}_1 = \frac{h}{\lambda_1} \hat{i}$ and $\vec{P}_2 = \frac{h}{\lambda_2} \hat{j}$.
Using the law of conservation of linear momentum,the momentum of the final particle $\vec{P}$ is the vector sum of the initial momenta:
$\vec{P} = \vec{P}_1 + \vec{P}_2 = \frac{h}{\lambda_1} \hat{i} + \frac{h}{\lambda_2} \hat{j}$.
The magnitude of the final momentum is:
$|\vec{P}| = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}$.
Since the de Broglie wavelength of the final particle is $\lambda = \frac{h}{|\vec{P}|}$,we have:
$\frac{h}{\lambda} = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}$.
Squaring both sides and dividing by $h^2$ gives:
$\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}$.

(D) Let the mass of nucleus $A$ be $2m$,so the mass of each nucleus $B$ and $C$ is $m$. Let the initial velocity of $A$ be $v_0$. The initial momentum is $P_A = (2m)v_0$.
According to the law of conservation of linear momentum,the final momentum must equal the initial momentum. Let $v$ be the velocity of $B$. Then the velocity of $C$ is $v/2$ in the opposite direction.
$P_f = m v - m(v/2) = m v / 2$.
Equating initial and final momentum: $2m v_0 = m v / 2$,which gives $v = 4v_0$.
Now,calculate the momenta of $B$ and $C$:
$P_B = m v = m(4v_0) = 4m v_0$.
$P_C = m(v/2) = m(2v_0) = 2m v_0$.
The de Broglie wavelength is given by $\lambda = h/P$.
Given $\lambda_A = h / (2m v_0)$.
$\lambda_B = h / P_B = h / (4m v_0) = \frac{1}{2} \times \frac{h}{2m v_0} = \frac{\lambda_A}{2}$.
$\lambda_C = h / P_C = h / (2m v_0) = \lambda_A$.
Thus,the wavelengths are $\lambda_B = \frac{\lambda_A}{2}$ and $\lambda_C = \lambda_A$.

(C) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.
Let the momenta of particles $x$ and $y$ be $\vec{p}_x$ and $\vec{p}_y$ respectively.
Since they move in opposite directions,we take one direction as positive and the other as negative.
The magnitude of momentum is related to the de-Broglie wavelength by $p = \frac{h}{\lambda}$.
Let $p_x = \frac{h}{\lambda_x}$ and $p_y = \frac{h}{\lambda_y}$.
The final momentum of particle $P$ is $p_f = |p_x - p_y|$.
Substituting the values,$p_f = |\frac{h}{\lambda_x} - \frac{h}{\lambda_y}| = h |\frac{1}{\lambda_x} - \frac{1}{\lambda_y}|$.
Since $p_f = \frac{h}{\lambda}$,we have $\frac{h}{\lambda} = h |\frac{\lambda_y - \lambda_x}{\lambda_x \lambda_y}|$.
Therefore,$\lambda = \frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$.

(A) Given that the de Broglie wavelength of the electron $\lambda_e$ is equal to the wavelength of the photon $\lambda_p$,so $\lambda_e = \lambda_p = \lambda$.
The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2$.
The energy of the photon is $E_p = \frac{hc}{\lambda_p}$.
From the de Broglie relation,$\lambda_e = \frac{h}{m_e v_e}$,which implies $m_e = \frac{h}{\lambda_e v_e}$.
Substituting $m_e$ into the kinetic energy expression: $K_e = \frac{1}{2} \left( \frac{h}{\lambda_e v_e} \right) v_e^2 = \frac{h v_e}{2 \lambda_e}$.
Now,the ratio of the kinetic energy of the electron to the energy of the photon is:
$\frac{K_e}{E_p} = \frac{h v_e / 2 \lambda_e}{hc / \lambda_p} = \frac{v_e}{2c}$.
Given $v_e = 1.5 \times 10^8 \, m/s$ and $c = 3 \times 10^8 \, m/s$:
$\frac{K_e}{E_p} = \frac{1.5 \times 10^8}{2 \times 3 \times 10^8} = \frac{1.5}{6} = \frac{1}{4}$.

(D) The initial momentum of the system is $P_{i} = 0$ because the neutron is slowly moving (momentum $\approx 0$) and the nucleus of mass $M$ is at rest.
According to the law of conservation of linear momentum,the final momentum $P_{f}$ must also be zero.
Let $P_{1}$ be the momentum of the nucleus with mass $m_{1}$ and $P_{2}$ be the momentum of the nucleus with mass $5m_{1}$.
Thus,$P_{1} + P_{2} = 0$,which implies $P_{1} = -P_{2}$.
Taking the magnitude,$|P_{1}| = |P_{2}| = P$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{P}$.
For the first nucleus,$\lambda_{1} = \frac{h}{P_{1}} = \lambda$.
For the second nucleus,$\lambda_{2} = \frac{h}{P_{2}} = \frac{h}{P_{1}} = \lambda$.
Therefore,the de-Broglie wavelength of the other nucleus is also $\lambda$.

(B) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $E$ by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.
From this,we see that $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_2}{E_1}}$.
Given $\lambda_1 = 10^{-10} \ m$ and $\lambda_2 = 0.5 \times 10^{-10} \ m$,we have $\frac{10^{-10}}{0.5 \times 10^{-10}} = \sqrt{\frac{E_2}{E_1}}$.
This simplifies to $2 = \sqrt{\frac{E_2}{E_1}}$,so $\frac{E_2}{E_1} = 4$,or $E_2 = 4E_1$.
The energy that must be added is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.
Thus,the added energy is thrice the initial energy.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.
Taking the derivative,we get $d\lambda = -\frac{h}{p^2} dp$.
Dividing by $\lambda$,we have $\frac{d\lambda}{\lambda} = -\frac{dp}{p}$.
Taking the magnitude,$\left| \frac{\Delta \lambda}{\lambda} \right| = \left| \frac{\Delta p}{p} \right|$.
Given that the change in wavelength is $0.25\%$,we have $\frac{\Delta \lambda}{\lambda} = \frac{0.25}{100} = \frac{1}{400}$.
Substituting the given change in momentum $\Delta p = p_0$,we get $\frac{p_0}{p} = \frac{1}{400}$.
Therefore,the original momentum $p = 400\,p_0$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.
Since the kinetic energy $E = 10^{-32} \ J$ is the same for both the electron and the proton,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
We know that the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,i.e.,$m_p > m_e$.
Therefore,$\sqrt{m_p} > \sqrt{m_e}$,which implies $\frac{1}{\sqrt{m_p}} < \frac{1}{\sqrt{m_e}}$.
Thus,the de-Broglie wavelength of the proton is less than that of the electron,i.e.,$\lambda_p < \lambda_e$.

(C) When a charged particle (charge $q$,mass $m$) enters a magnetic field $B$ perpendicularly,the radius $r$ of its circular path is given by:
$r = \frac{mv}{qB} \Rightarrow mv = qBr$
The de-Broglie wavelength $\lambda$ is given by:
$\lambda = \frac{h}{mv}$
Substituting $mv = qBr$ into the wavelength formula:
$\lambda = \frac{h}{qBr}$
Given that the radii $r$ and the magnetic field $B$ are the same for both particles:
$\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{q_{p}}{q_{\alpha}}$
For an $\alpha$-particle,$q_{\alpha} = 2e$,and for a proton,$q_{p} = e$:
$\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{e}{2e} = \frac{1}{2}$

(C) The de Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2meV}}$
where $h$ is Planck's constant and $e$ is the charge of the particle.
Since both the proton and the electron are accelerated by the same potential difference $V$ and have the same magnitude of charge $e$,the wavelength is inversely proportional to the square root of the mass of the particle:
$\lambda \propto \frac{1}{\sqrt{m}}$
We know that the mass of an electron $(m_e)$ is significantly less than the mass of a proton $(m_p)$,i.e.,$m_e < m_p$.
Therefore,$\frac{1}{\sqrt{m_e}} > \frac{1}{\sqrt{m_p}}$,which implies $\lambda_e > \lambda_p$.

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Since the particle and the electron have the same wavelength,their momenta must be equal.
$m_p v_p = m_e v_e$
Here,$m_p = 1\, mg = 10^{-6}\, kg$,$m_e = 9.1 \times 10^{-31}\, kg$,and $v_e = 3 \times 10^6\, m\,s^{-1}$.
Substituting the values:
$v_p = \frac{m_e v_e}{m_p} = \frac{9.1 \times 10^{-31} \times 3 \times 10^6}{10^{-6}}$
$v_p = 27.3 \times 10^{-25} \times 10^6 = 27.3 \times 10^{-19} = 2.73 \times 10^{-18}\, m\,s^{-1}$.
Rounding to the given options,the velocity is $2.7 \times 10^{-18}\, m\,s^{-1}$.

(A) 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 energy $E$ is the same for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}}$.
We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton $(m_\alpha \approx 4m_p)$.
Substituting this into the ratio: $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(C) The kinetic energy $E$ of a particle of mass $m$ moving with velocity $v$ is given by $E = \frac{1}{2}mv^2$.
Multiplying both sides by $2m$,we get $2mE = m^2v^2$.
Taking the square root of both sides,we find the momentum $p = mv = \sqrt{2mE}$.
According to the De Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.
Substituting the value of $p$,we get $\lambda = \frac{h}{\sqrt{2mE}}$.

(D) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is given by $K = qV$.
Since $K = p^2 / 2m$,where $p$ is the momentum and $m$ is the mass,we have $p = \sqrt{2mqV}$.
For an electron,$q_e = e$ and mass is $m_e$. Thus,$p_e = \sqrt{2m_e eV}$.
For an alpha particle,$q_\alpha = 2e$ and mass is $m_\alpha$. Thus,$p_\alpha = \sqrt{2m_\alpha (2e) V} = \sqrt{4m_\alpha eV}$.
The ratio of momenta is $\frac{p_e}{p_\alpha} = \frac{\sqrt{2m_e eV}}{\sqrt{4m_\alpha eV}} = \sqrt{\frac{2m_e}{4m_\alpha}} = \sqrt{\frac{m_e}{2m_\alpha}}$.

(D) The electron enters the region between the plates with an initial kinetic energy of $200 \, eV$.
The electric field between the plates is directed from the positive plate to the negative plate. Since the electron is negatively charged,it experiences a force opposite to the direction of the electric field,i.e.,it is decelerated.
The potential difference between the plates is $V = 100 \, V$,so the work done against the electric field as the electron moves from the first plate to the second plate is $W = qV = 100 \, eV$.
The final kinetic energy of the electron when it emerges from the second plate is $K_f = K_i - W = 200 \, eV - 100 \, eV = 100 \, eV$.
The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A}$,where $V$ is the kinetic energy in electron-volts $(eV)$.
Substituting $V = 100 \, eV$,we get $\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \, \mathring{A}$.
Rounding to two decimal places,we get $\lambda \approx 1.23 \, \mathring{A}$.
Since $1.23 \, \mathring{A}$ is not among the given options,the correct choice is 'None of these'.

(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
For an $\alpha$-particle,the mass $m_{\alpha} = 4u$ and velocity $v_{\alpha} = v$. Thus,$\lambda_{\alpha} = \frac{h}{4v}$.
For a deuteron,the mass $m_{D} = 2u$ and velocity $v_{D} = 2v$. Thus,$\lambda_{D} = \frac{h}{2 \times 2v} = \frac{h}{4v}$.
Taking the ratio,$\frac{\lambda_{\alpha}}{\lambda_{D}} = \frac{h/4v}{h/4v} = 1:1$.

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For a gas molecule at temperature $T$,the root mean square speed is $v = \sqrt{\frac{3kT}{m}}$,where $k$ is the Boltzmann constant and $m$ is the mass of the molecule.
Substituting $v$ into the wavelength formula: $\lambda = \frac{h}{m\sqrt{\frac{3kT}{m}}} = \frac{h}{\sqrt{3mkT}}$.
For hydrogen $(H_2)$ at $T_1 = 27\,^oC = 300\,K$ and helium $(He)$ at $T_2 = 127\,^oC = 400\,K$,the masses are $m_H = 2\,amu$ and $m_{He} = 4\,amu$.
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 de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$
Since both the proton and electron are accelerated through the same potential $V$ and have the same magnitude of charge $q = e$,we have:
$\lambda \propto \frac{1}{\sqrt{m}}$
Therefore,the ratio of the wavelengths is:
$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energy.
For a proton,$\lambda_p = \frac{h}{\sqrt{2m_p K}}$.
For an $\alpha$-particle,the mass is $m_{\alpha} = 4m_p$. Given the kinetic energies are equal $(K_p = K_{\alpha} = K)$,we have $\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)K}} = \frac{h}{2\sqrt{2m_p K}}$.
Comparing the two,we get $\lambda_{\alpha} = \frac{\lambda_p}{2}$,which implies $\lambda_p = 2\lambda_{\alpha}$.

(C) The resolving power of a microscope is inversely proportional to the wavelength of the radiation used,given by the relation $RP \propto \frac{1}{\lambda}$.
Visible light has a wavelength in the range of $4000 \, \mathring{A}$ to $7000 \, \mathring{A}$.
According to the de Broglie hypothesis,the wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For an electron accelerated through a potential difference $V$,the wavelength is approximately $\lambda \approx \frac{12.27}{\sqrt{V}} \, \mathring{A}$.
Even for a small accelerating voltage,the wavelength of an electron is significantly smaller than the wavelength of visible light.
Since $\lambda_{\text{electron}} < \lambda_{\text{light}}$,the resolving power of an electron microscope is much higher than that of an optical microscope.

(B) The root mean square velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3kT}{m}}$.
The de-Broglie wavelength is $\lambda = \frac{h}{mv_{rms}} = \frac{h}{m} \sqrt{\frac{m}{3kT}} = \frac{h}{\sqrt{3mkT}}$.
For hydrogen $(H_2)$,$m_H = 2$ units and $T_H = 27 + 273 = 300\text{ K}$.
For helium $(He)$,$m_{He} = 4$ units and $T_{He} = 127 + 273 = 400\text{ K}$.
The ratio is $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{m_{He} T_{He}}{m_H T_H}}$.
Substituting the values: $\frac{\lambda_H}{\lambda_{He}} = \sqrt{\frac{4 \times 400}{2 \times 300}} = \sqrt{\frac{1600}{600}} = \sqrt{\frac{16}{6}} = \sqrt{\frac{8}{3}}$.

(B) The de Broglie wavelength $\lambda$ for an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \; \mathring{A}$.
Given $V = 10,000 \; V = 10^4 \; V$.
Substituting the value of $V$ in the formula:
$\lambda = \frac{12.27}{\sqrt{10^4}} \; \mathring{A} = \frac{12.27}{100} \; \mathring{A} = 0.1227 \; \mathring{A}$.
Since $1 \; \mathring{A} = 10^{-10} \; m$,we have $\lambda = 0.1227 \times 10^{-10} \; m = 12.27 \times 10^{-12} \; m$.
Rounding to the nearest option,the value is approximately $12.2 \times 10^{-12} \; m$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E_k$ is the kinetic energy.
Since both particles have the same kinetic energy $E_k$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}}$.
We know that the mass of an $\alpha$-particle is approximately $4$ times the mass of a proton,i.e.,$m_{\alpha} = 4m_p$.
Substituting this value,we get $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = 2$.
Thus,the ratio $\lambda_p : \lambda_{\alpha}$ is $2:1$.

(B) The energy of a photon is given by $E = \frac{hc}{\lambda_{\text{photon}}}$,which implies $\lambda_{\text{photon}} = \frac{hc}{E}$.
The de-Broglie wavelength of an electron with energy $E$ is given by $\lambda_{\text{electron}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_{\text{electron}}}{\lambda_{\text{photon}}} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc}$.
Simplifying the expression:
$\frac{\lambda_{\text{electron}}}{\lambda_{\text{photon}}} = \frac{1}{c} \times \frac{E}{\sqrt{2mE}} = \frac{1}{c} \times \sqrt{\frac{E^2}{2mE}} = \frac{1}{c} \left(\frac{E}{2m}\right)^{1/2}$.

(C) The initial velocity is $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$. The initial speed is $v = \sqrt{v_{0}^{2} + v_{0}^{2}} = v_{0} \sqrt{2}$.
The initial de-Broglie wavelength is $\lambda_{0} = \frac{h}{m v_{0} \sqrt{2}} \implies h = \lambda_{0} m v_{0} \sqrt{2}$.
The electric field is $\overrightarrow{E} = -E_{0} \hat{k}$. The force on the electron is $\overrightarrow{F} = -e \overrightarrow{E} = e E_{0} \hat{k}$.
The acceleration is $\overrightarrow{a} = \frac{e E_{0}}{m} \hat{k}$.
At time $t$,the velocity is $\overrightarrow{v}(t) = v_{0} \hat{i} + v_{0} \hat{j} + \frac{e E_{0} t}{m} \hat{k}$.
The speed at time $t$ is $v(t) = \sqrt{v_{0}^{2} + v_{0}^{2} + \left(\frac{e E_{0} t}{m}\right)^{2}} = \sqrt{2 v_{0}^{2} + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}$.
The de-Broglie wavelength at time $t$ is $\lambda = \frac{h}{m v(t)} = \frac{\lambda_{0} m v_{0} \sqrt{2}}{m \sqrt{2 v_{0}^{2} + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}} = \frac{\lambda_{0} v_{0} \sqrt{2}}{\sqrt{2 v_{0}^{2} + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}} = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{2 m^{2} v_{0}^{2}}}}$.

(A) The acceleration of the electron is $a = \frac{|e|E}{m}$.
Since the electron starts from rest $(u = 0)$,its velocity at time $t$ is $v = at = \frac{|e|E}{m}t$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv} = \frac{h}{m(\frac{|e|E}{m}t)} = \frac{h}{|e|Et}$.
To find the rate of change of wavelength,we differentiate $\lambda$ with respect to $t$:
$\frac{d\lambda}{dt} = \frac{d}{dt} (\frac{h}{|e|Et}) = \frac{h}{|e|E} \cdot \frac{d}{dt}(t^{-1}) = \frac{h}{|e|E} (-t^{-2}) = -\frac{h}{|e|Et^2}$.

(C) The de Broglie wavelength $\lambda$ of a particle with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.
When the energy is increased by $\Delta E$,the new kinetic energy becomes $E' = E + \Delta E$,and the new wavelength is $\lambda' = \frac{\lambda}{2}$.
Using the formula for the new wavelength: $\lambda' = \frac{h}{\sqrt{2m(E + \Delta E)}}$.
Since $\lambda' = \frac{\lambda}{2}$,we have $\frac{h}{\sqrt{2m(E + \Delta E)}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2mE}}$.
Squaring both sides: $\frac{1}{2m(E + \Delta E)} = \frac{1}{4} \cdot \frac{1}{2mE}$.
Simplifying the equation: $\frac{1}{E + \Delta E} = \frac{1}{4E}$.
Therefore,$4E = E + \Delta E$,which gives $\Delta E = 3E$.

(N/A) Matter waves,also known as de Broglie waves,are the waves associated with moving particles of matter (such as electrons,protons,or neutrons). According to the de Broglie hypothesis,any moving particle with momentum $p$ has an associated wavelength $\lambda$ given by $\lambda = h/p$,where $h$ is Planck's constant.
Devices that employ the use of matter waves include:
$1$. Electron Microscope: It uses the wave nature of electrons to achieve much higher resolution than optical microscopes.
$2$. Electron Diffraction Apparatus: Used to study the crystal structure of materials by observing the diffraction patterns of electrons.

(N/A) For the electron:
Mass $m = 9.11 \times 10^{-31} \; kg$,speed $v = 5.4 \times 10^{6} \; m/s$.
Momentum $p = mv = (9.11 \times 10^{-31} \; kg) \times (5.4 \times 10^{6} \; m/s) = 4.92 \times 10^{-24} \; kg \cdot m/s$.
de Broglie wavelength $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \; J \cdot s}{4.92 \times 10^{-24} \; kg \cdot m/s} \approx 1.35 \times 10^{-10} \; m = 0.135 \; nm$.
$(b)$ For the ball:
Mass $m' = 0.150 \; kg$,speed $v' = 30.0 \; m/s$.
Momentum $p' = m'v' = (0.150 \; kg) \times (30.0 \; m/s) = 4.50 \; kg \cdot m/s$.
de Broglie wavelength $\lambda' = \frac{h}{p'} = \frac{6.63 \times 10^{-34} \; J \cdot s}{4.50 \; kg \cdot m/s} \approx 1.47 \times 10^{-34} \; m$.
The de Broglie wavelength of the electron is comparable to $X$-ray wavelengths,whereas for the ball,it is extremely small (about $10^{-19}$ times the size of a proton),making it beyond experimental measurement.

(C) The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = h / p$.
Since kinetic energy $K = p^2 / (2m)$,we can write momentum $p = \sqrt{2mK}$.
Substituting this into the wavelength formula,we get $\lambda = h / \sqrt{2mK}$.
For a constant kinetic energy $K$,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto 1 / \sqrt{m}$.
Comparing the masses: $m_{\alpha} > m_p > m_e$.
Since the $\alpha$-particle has the largest mass among the three,it will have the smallest de Broglie wavelength.

(N/A) The de Broglie wavelength of a particle with mass $m$ and velocity $v$ is given by $\lambda = \frac{h}{mv}$.
For an electron,the wavelength is $\lambda_e = \frac{h}{m_e v_e}$.
Given that the particle's velocity $v = 3v_e$ and the ratio of wavelengths $\frac{\lambda}{\lambda_e} = 1.813 \times 10^{-4}$.
Using the ratio: $\frac{\lambda}{\lambda_e} = \frac{h/mv}{h/m_e v_e} = \frac{m_e v_e}{mv} = 1.813 \times 10^{-4}$.
Rearranging for the mass of the particle $m$: $m = m_e \left( \frac{v_e}{v} \right) \left( \frac{\lambda_e}{\lambda} \right)$.
Substituting the values $m_e = 9.11 \times 10^{-31} \ kg$,$\frac{v_e}{v} = \frac{1}{3}$,and $\frac{\lambda_e}{\lambda} = \frac{1}{1.813 \times 10^{-4}}$:
$m = (9.11 \times 10^{-31} \ kg) \times \left( \frac{1}{3} \right) \times \left( \frac{1}{1.813 \times 10^{-4}} \right) \approx 1.675 \times 10^{-27} \ kg$.
This mass corresponds to the mass of a proton or a neutron.

(C) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Given:
$h = 6.6 \times 10^{-34}\; Js$
$m = 9.1 \times 10^{-31}\; kg$
$E = 100\; eV = 100 \times 1.6 \times 10^{-19}\; J = 1.6 \times 10^{-17}\; J$.
Substituting these values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{29.12 \times 10^{-48}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{5.396 \times 10^{-24}}$
$\lambda \approx 1.22 \times 10^{-10}\; m$.
Since $1\; \mathring{A} = 10^{-10}\; m$,the wavelength is approximately $1.2\; \mathring{A}$.

(N/A) The momentum $p$ of an electron accelerated through a potential difference $V$ is given by the formula $p = \sqrt{2meV}$.
Substituting the values: $m = 9.11 \times 10^{-31} \; kg$,$e = 1.602 \times 10^{-19} \; C$,and $V = 56 \; V$.
$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times 56} \approx 4.04 \times 10^{-24} \; kg \cdot m/s$.
$(b)$ The de Broglie wavelength $\lambda$ is given by $\lambda = h/p$.
Using $h = 6.626 \times 10^{-34} \; J \cdot s$ and $p = 4.04 \times 10^{-24} \; kg \cdot m/s$:
$\lambda = (6.626 \times 10^{-34}) / (4.04 \times 10^{-24}) \approx 1.64 \times 10^{-10} \; m = 0.164 \; nm$.

$(a)$ The momentum $p$ is given by $p = \sqrt{2mK}$. Given $K = 120 \ eV = 120 \times 1.6 \times 10^{-19} \ J = 1.92 \times 10^{-17} \ J$ and $m = 9.11 \times 10^{-31} \ kg$. Thus,$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.92 \times 10^{-17}} \approx 5.91 \times 10^{-24} \ kg \ m/s$.
$(b)$ The speed $v$ is given by $K = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 6.5 \times 10^6 \ m/s$.
$(c)$ The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.91 \times 10^{-24}} \approx 1.12 \times 10^{-10} \ m = 0.112 \ nm$.

(N/A) Given,wavelength $\lambda = 589 \; nm = 589 \times 10^{-9} \; m$.
The momentum $p$ associated with this wavelength is given by $p = \frac{h}{\lambda}$,where $h = 6.626 \times 10^{-34} \; J \cdot s$.
$p = \frac{6.626 \times 10^{-34}}{589 \times 10^{-9}} \approx 1.125 \times 10^{-27} \; kg \cdot m/s$.
$(a)$ For an electron,mass $m_e = 9.11 \times 10^{-31} \; kg$. The kinetic energy $K_e = \frac{p^2}{2m_e}$.
$K_e = \frac{(1.125 \times 10^{-27})^2}{2 \times 9.11 \times 10^{-31}} \approx 6.95 \times 10^{-25} \; J$.
$(b)$ For a neutron,mass $m_n = 1.675 \times 10^{-27} \; kg$. The kinetic energy $K_n = \frac{p^2}{2m_n}$.
$K_n = \frac{(1.125 \times 10^{-27})^2}{2 \times 1.675 \times 10^{-27}} \approx 3.78 \times 10^{-28} \; J$.

(N/A) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $h = 6.626 \times 10^{-34} \; J \cdot s$.
$(a)$ Given $m = 0.040 \; kg$,$v = 1.0 \; km/s = 1000 \; m/s$.
$\lambda = \frac{6.626 \times 10^{-34}}{0.040 \times 1000} = \frac{6.626 \times 10^{-34}}{40} \approx 1.66 \times 10^{-35} \; m$.
$(b)$ Given $m = 0.060 \; kg$,$v = 1.0 \; m/s$.
$\lambda = \frac{6.626 \times 10^{-34}}{0.060 \times 1.0} \approx 1.10 \times 10^{-32} \; m$.
$(c)$ Given $m = 1.0 \times 10^{-9} \; kg$,$v = 2.2 \; m/s$.
$\lambda = \frac{6.626 \times 10^{-34}}{1.0 \times 10^{-9} \times 2.2} \approx 3.01 \times 10^{-25} \; m$.