Matter Waves and de Broglie Wavelength Questions in English

(B) The de Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum and $h$ is Planck's constant.
The kinetic energy $K$ of a particle is related to its momentum $p$ by $K = \frac{p^2}{2m}$,where $m$ is the mass of the particle.
From this,we can express the momentum as $p = \sqrt{2mK}$. Substituting this into the de Broglie wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.
Since the mass $m$ of the particle remains constant,the relationship between wavelength and kinetic energy is $\lambda \propto \frac{1}{\sqrt{K}}$.
Given $\lambda = 2000 \ \mathring{A}$ for $K = 1 \ \text{eV}$,and we need to find $\lambda'$ for $K' = 1 \ \text{MeV} = 10^6 \ \text{eV}$.
Using the ratio: $\frac{\lambda'}{\lambda} = \sqrt{\frac{K}{K'}} = \sqrt{\frac{1 \ \text{eV}}{10^6 \ \text{eV}}} = \sqrt{\frac{1}{10^6}} = \frac{1}{10^3}$.
Therefore,$\lambda' = \frac{\lambda}{10^3} = \frac{2000}{1000} \ \mathring{A} = 2 \ \mathring{A}$.

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial wavelength be $\lambda_1 = \lambda$ at potential $V_1 = V$.
Let the new wavelength be $\lambda_2$ at potential $V_2 = 9V$.
Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$.
Substituting the values: $\frac{\lambda_2}{\lambda} = \sqrt{\frac{V}{9V}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Therefore,$\lambda_2 = \frac{\lambda}{3}$.

(C) The energy of the incident photon is $E = \frac{hc}{\lambda}$.
Since the work function is negligible,the kinetic energy $K$ of the emitted photoelectron is equal to the energy of the incident photon: $K = \frac{hc}{\lambda}$.
The de-Broglie wavelength $\lambda_1$ of the electron is given by $\lambda_1 = \frac{h}{p}$,where $p$ is the momentum of the electron.
We know that $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.
Substituting $K = \frac{hc}{\lambda}$,we get $p = \sqrt{2m \cdot \frac{hc}{\lambda}}$.
Thus,$\lambda_1 = \frac{h}{\sqrt{\frac{2mhc}{\lambda}}}$.
Squaring both sides,we get $\lambda_1^2 = \frac{h^2}{\frac{2mhc}{\lambda}} = \frac{h^2 \lambda}{2mhc} = \frac{h \lambda}{2mc}$.
Rearranging the terms,we find $\frac{\lambda}{\lambda_1^2} = \frac{2mc}{h}$.
Therefore,the ratio $\lambda : \lambda_1^2$ is $2mc : h$.

(A) 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 (impulse) 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 (impulse) of the particle.
Thus,option $A$ is correct.

(C) The de-Broglie wavelength $\lambda$ of a neutron is given by $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $E_k$ is the kinetic energy.
For a neutron in thermal equilibrium at temperature $T$,the average kinetic energy is $E_k = \frac{3}{2} k_B T$.
Thus,$\lambda = \frac{h}{\sqrt{2m(\frac{3}{2} k_B T)}} = \frac{h}{\sqrt{3mk_B T}}$.
This implies $\lambda \propto \frac{1}{\sqrt{T}}$.
Given $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$ and $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.
Therefore,$\frac{\lambda_2}{\lambda_0} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
So,$\lambda_2 = \frac{\lambda_0}{2}$.

(B) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.
From this,we have $E = \frac{h^2}{2m\lambda^2}$,which means $E \propto \frac{1}{\lambda^2}$.
Let the initial energy be $E_1$ corresponding to wavelength $\lambda$. So,$E_1 = \frac{k}{\lambda^2}$ (where $k = \frac{h^2}{2m}$).
The final wavelength is $\lambda_2 = \frac{\lambda}{2}$. The final energy $E_2$ is $E_2 = \frac{k}{(\lambda/2)^2} = \frac{4k}{\lambda^2} = 4E_1$.
The energy to be added is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.
Given that $\Delta E = nE_1$,we find $n = 3$.

(D) The de-Broglie wavelength $\lambda$ of an electron accelerated by a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let $\lambda_1$ be the initial wavelength at voltage $V_1 = 10 \ kV$ and $\lambda_2$ be the final wavelength at voltage $V_2 = 20 \ kV$.
Then,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$.
Substituting the given values: $\frac{\lambda_2}{\lambda} = \sqrt{\frac{10 \ kV}{20 \ kV}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\lambda_2 = \frac{\lambda}{\sqrt{2}}$.

(A) The force on the electron is $F = eE$. The acceleration is $a = \frac{eE}{m}$.
Since the electron starts from rest,its velocity at time $t$ is $v = at = \frac{eEt}{m}$.
The momentum of the electron is $p = mv = m(\frac{eEt}{m}) = eEt$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{eEt}$.
To find the rate of change of wavelength,we differentiate $\lambda$ with respect to time $t$:
$\frac{d\lambda}{dt} = \frac{d}{dt}(\frac{h}{eEt}) = \frac{h}{eE} \frac{d}{dt}(t^{-1}) = \frac{h}{eE} (-t^{-2}) = -\frac{h}{eEt^2}$.

(C) $1$. Energy of a photon $(E_p)$ is given by $E_p = \frac{hc}{\lambda}$.
$2$. De-Broglie wavelength of an electron is $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron. Thus,$p = \frac{h}{\lambda}$.
$3$. Kinetic energy of the electron $(K_e)$ is given by $K_e = \frac{p^2}{2m}$.
$4$. Substituting $p = \frac{h}{\lambda}$ into the kinetic energy formula: $K_e = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$.
$5$. The ratio of the kinetic energy of the electron to the energy of the photon is $\frac{K_e}{E_p} = \frac{h^2 / (2m\lambda^2)}{hc / \lambda}$.
$6$. Simplifying the expression: $\frac{K_e}{E_p} = \frac{h^2}{2m\lambda^2} \times \frac{\lambda}{hc} = \frac{h}{2mc\lambda}$.

(B) The de-Broglie wavelength of an electron is given by $\lambda_e = h / p_e = h / (m_e v)$.
Given $v = c / 100$,so $\lambda_e = h / (m_e c / 100) = 100h / (m_e c)$.
The kinetic energy of the electron is $E_e = (1/2) m_e v^2 = (1/2) m_e (c / 100)^2 = m_e c^2 / 20000$.
The de-Broglie wavelength of a photon is $\lambda_p = h / p_p = hc / E_p$,which implies $E_p = hc / \lambda_p$.
Given $\lambda_p = 2 \lambda_e$,we substitute $\lambda_e$:
$\lambda_p = 2 \times (100h / (m_e c)) = 200h / (m_e c)$.
Now,calculate $E_p = hc / (200h / (m_e c)) = m_e c^2 / 200$.
Finally,the ratio $E_p / E_e = (m_e c^2 / 200) / (m_e c^2 / 20000) = 20000 / 200 = 100 = 10^2$.

(B) For a photon,the energy is $E_p = h\nu = hc / \lambda_p$.
For an electron,the de-Broglie wavelength is $\lambda_e = h / p_e$,so $p_e = h / \lambda_e$.
The kinetic energy of the electron is $E_e = p_e^2 / (2m) = h^2 / (2m \lambda_e^2)$.
Given $\lambda_p = 2\lambda_e$,we have $\lambda_e = \lambda_p / 2$.
Substituting this into the electron energy expression: $E_e = h^2 / (2m (\lambda_p / 2)^2) = 2h^2 / (m \lambda_p^2)$.
Now,find the ratio $E_e / E_p$:
$E_e / E_p = [2h^2 / (m \lambda_p^2)] / [hc / \lambda_p] = 2h / (mc \lambda_p)$.
Since $\lambda_e = h / (m v_e)$,we have $\lambda_p = 2 \lambda_e = 2h / (m v_e)$.
Substitute $\lambda_p$ into the ratio: $E_e / E_p = 2h / (mc \cdot (2h / (m v_e))) = v_e / c$.
Given $v_e = C / 100$,the ratio is $E_e / E_p = (C / 100) / C = 1 / 100 = 1 \times 10^{-2}$.

(B) The kinetic energy $E$ is related to momentum $p$ by the formula $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.
According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$.
If the kinetic energy is increased by $2$ times,the new kinetic energy $E'$ becomes $E + 2E = 3E$.
The new wavelength $\lambda'$ is $\lambda' = \frac{h}{\sqrt{2m(3E)}} = \frac{1}{\sqrt{3}} \lambda$.
Therefore,the wavelength changes by a factor of $\frac{1}{\sqrt{3}}$.

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the relation: $\lambda = \frac{h}{\sqrt{2meV}}$.
This implies that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial potential be $V_1$ and the final potential be $V_2 = 4V_1$.
The ratio of the wavelengths is given by: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$.
Substituting the value of $V_2$: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{4V_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$\lambda_2 = \frac{\lambda_1}{2}$.
Thus,the de-Broglie wavelength decreases to half of its initial value.

(C) The de Broglie wavelength $(\lambda)$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial potential be $V_1 = V$ and the final potential be $V_2 = 2V$.
The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.
Therefore,the new wavelength $\lambda_2 = \frac{1}{\sqrt{2}} \lambda_1$.
Thus,the de Broglie wavelength is decreased by a factor of $\frac{1}{\sqrt{2}}$.

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the relation:
$\lambda = \frac{h}{\sqrt{2meV}} \implies \lambda \propto \frac{1}{\sqrt{V}}$
Given initial potential $V_1 = 16 \ kV$ and final potential $V_2 = 64 \ kV$.
Using the ratio formula:
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{16}{64}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Therefore,$\lambda_2 = \frac{\lambda_1}{2}$.
Thus,the de-Broglie wavelength becomes half of its initial value.

(D) For a proton,the kinetic energy $E$ is related to its momentum $p$ by the equation $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.
From this,the momentum is $p = \sqrt{2mE}$.
The de-Broglie wavelength of the proton is $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon,the energy $E$ is related to its wavelength $\lambda_2$ by the equation $E = \frac{hc}{\lambda_2}$,where $c$ is the speed of light.
Thus,the wavelength of the photon is $\lambda_2 = \frac{hc}{E}$.
Now,taking the ratio of the wavelengths:
$\frac{\lambda_1}{\lambda_2} = \left( \frac{h}{\sqrt{2mE}} \right) \times \left( \frac{E}{hc} \right) = \frac{1}{c} \sqrt{\frac{E}{2m}}$.
This simplifies to $\frac{\lambda_1}{\lambda_2} = \left( \frac{1}{c\sqrt{2m}} \right) E^{1/2}$.
Comparing this with $\frac{\lambda_1}{\lambda_2} \propto E^n$,we find that $n = 1/2 = 0.5$.

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.228}{\sqrt{V}} \text{ nm}$.
From this relation,we see that $\lambda \propto \frac{1}{\sqrt{V}}$,which implies $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$ or $\frac{V_1}{V_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^2$.
Given that the wavelength increases by $50 \%$,the new wavelength $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$.
Substituting this into the ratio:
$\frac{V_1}{V_2} = \left(\frac{1.5\lambda_1}{\lambda_1}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial potential difference be $V_1 = V$ and the final potential difference be $V_2 = 2V$.
The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.
Therefore,the de-Broglie wavelength decreases by a factor of $\frac{1}{\sqrt{2}}$.

(D) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial wavelength be $\lambda_1 = 4\lambda$ at potential $V_1 = V$.
Let the final wavelength be $\lambda_2$ at potential $V_2 = 4V$.
Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$,we get:
$\frac{\lambda_2}{4\lambda} = \sqrt{\frac{V}{4V}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$\lambda_2 = 4\lambda \times \frac{1}{2} = 2\lambda$.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.
Since $p = \sqrt{2mqV}$,we have $\lambda = \frac{h}{\sqrt{2mqV}}$.
Rearranging the terms,we get $\lambda = \left( \frac{h}{\sqrt{2mq}} \right) \left( \frac{1}{\sqrt{V}} \right)$.
Comparing this with the equation of a straight line $y = mx$,where $y = \lambda$ and $x = \frac{1}{\sqrt{V}}$,the slope of the graph is given by $\text{Slope} = \frac{h}{\sqrt{2mq}}$.
Since $h$ and $q$ are constants,the slope is inversely proportional to the square root of the mass: $\text{Slope} \propto \frac{1}{\sqrt{m}}$.
Therefore,a smaller slope corresponds to a larger mass.
Looking at the graph,the line corresponding to $m_1$ has the smallest slope.
Thus,$m_1$ represents the particle with the largest mass.

(D) The de-Broglie wavelength $(\lambda)$ of a particle is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the particle.
We know that the relationship between kinetic energy $(E)$ and momentum $(p)$ is $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.
Substituting this into the de-Broglie wavelength equation,we get $\lambda = \frac{h}{\sqrt{2mE}}$.
Since $h$ and $m$ (mass of the particle) are constants,we can conclude that $\lambda \propto \frac{1}{\sqrt{E}}$,which is equivalent to $\lambda \propto E^{-\frac{1}{2}}$.

(B) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p}$.
For an electron with energy $E$ and mass $m$,the momentum $p_e = \sqrt{2mE}$. Thus,$\lambda_e = \frac{h}{\sqrt{2mE}}$.
For a photon with energy $E$,the momentum $p_p = \frac{E}{c}$. Thus,$\lambda_p = \frac{h}{p_p} = \frac{hc}{E}$.
The ratio of the wavelengths is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \frac{E}{\sqrt{2mE}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{c} \left[\frac{E}{2m}\right]^{1/2}$.

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial potential difference be $V_1 = V$ and the final potential difference be $V_2 = 2V$.
The ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.
Therefore,the de-Broglie wavelength decreases by a factor of $\frac{1}{\sqrt{2}}$.

(B) The de-Broglie wavelength of a charged particle accelerated through a potential difference '$V$' is given by:
$\lambda = \frac{h}{\sqrt{2mqV}}$
For an electron of mass '$m$' and charge '$q$':
$\lambda = \frac{h}{\sqrt{2mqV}}$ ---$(1)$
For a proton of mass '$M$' and charge '$q$' (since the magnitude of charge on an electron and a proton is the same):
$\lambda_p = \frac{h}{\sqrt{2Mq(9V)}}$ ---$(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\lambda}{\lambda_p} = \frac{\frac{h}{\sqrt{2mqV}}}{\frac{h}{\sqrt{2Mq(9V)}}} = \sqrt{\frac{2Mq(9V)}{2mqV}} = \sqrt{\frac{9M}{m}} = 3\sqrt{\frac{M}{m}}$
Therefore,the wavelength associated with the proton is:
$\lambda_p = \frac{\lambda}{3} \sqrt{\frac{m}{M}}$

(D) The relation between kinetic energy $K$ and momentum $P$ is given by $P = \sqrt{2mK}$.
If the kinetic energy is tripled,the new kinetic energy $K' = 3K$.
The new momentum $P'$ is given by $P' = \sqrt{2m(3K)} = \sqrt{3} \sqrt{2mK} = \sqrt{3}P$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = h/P$.
Therefore,the new wavelength $\lambda'$ is $\lambda' = h/P' = h/(\sqrt{3}P) = \lambda / \sqrt{3}$.
Thus,the wavelength changes by a factor of $1/\sqrt{3}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Since kinetic energy $K = \frac{p^2}{2m}$,we can express momentum as $p = \sqrt{2mK}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.
This shows that $\lambda \propto \frac{1}{\sqrt{K}}$.
If the kinetic energy $K$ is doubled to $K' = 2K$,the new wavelength $\lambda'$ becomes $\lambda' = \frac{h}{\sqrt{2m(2K)}} = \frac{1}{\sqrt{2}} \cdot \frac{h}{\sqrt{2mK}} = \frac{\lambda}{\sqrt{2}}$.
Therefore,the wavelength changes by a factor of $\frac{1}{\sqrt{2}}$.

(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.
Since the kinetic energy $K = \frac{p^2}{2m} = qV$,the momentum is $p = \sqrt{2mqV}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mqV}} = \left( \frac{h}{\sqrt{2mq}} \right) \left( \frac{1}{\sqrt{V}} \right)$.
Comparing this with the equation of a straight line $y = mx$,where $y = \lambda$ and $x = \frac{1}{\sqrt{V}}$,the slope is given by $\text{slope} = \frac{h}{\sqrt{2mq}}$.
Since $h$ and $q$ are constants,the slope is inversely proportional to the square root of the mass,i.e.,$\text{slope} \propto \frac{1}{\sqrt{m}}$.
Therefore,a larger slope corresponds to a smaller mass.
From the graph,the line corresponding to $m_4$ has the largest slope.
Thus,$m_4$ represents the particle with the smallest mass.

(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 express energy as $E = \frac{h^2}{2m\lambda^2}$.
Let the initial wavelength be $\lambda_1 = 1 \,nm$ and the final wavelength be $\lambda_2 = 0.5 \,nm$.
The initial energy is $E_1 = \frac{h^2}{2m\lambda_1^2}$.
The final energy is $E_2 = \frac{h^2}{2m\lambda_2^2} = \frac{h^2}{2m(0.5\lambda_1)^2} = \frac{h^2}{2m(0.25\lambda_1^2)} = 4E_1$.
The additional energy required is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.
Thus, the additional energy to be supplied is three times its initial energy.

(C) In the region $0 \leq x \leq 1$,the potential energy of the particle is $U = E$. The total energy is $E_{total} = nE$.
Since $E_{total} = K.E. + P.E.$,the kinetic energy $K_1 = nE - E = (n-1)E$.
The momentum $p_1 = \sqrt{2mK_1} = \sqrt{2m(n-1)E}$.
The de-Broglie wavelength is $\lambda_1 = \frac{h}{p_1} = \frac{h}{\sqrt{2m(n-1)E}}$.
In the region $x > 1$,the potential energy is $U = 0$.
Thus,the kinetic energy $K_2 = nE - 0 = nE$.
The momentum $p_2 = \sqrt{2mK_2} = \sqrt{2mnE}$.
The de-Broglie wavelength is $\lambda_2 = \frac{h}{p_2} = \frac{h}{\sqrt{2mnE}}$.
Taking the ratio,$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2m(n-1)E}}{h / \sqrt{2mnE}} = \sqrt{\frac{2mnE}{2m(n-1)E}} = \sqrt{\frac{n}{n-1}}$.

(B) The de-Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
Here,$h$ is Planck's constant,$m$ is the mass of the particle,and $q$ is the charge of the particle.
For a proton,let mass be $m_p = m$ and charge be $q_p = e$.
For an alpha particle,mass $m_{\alpha} = 4m$ and charge $q_{\alpha} = 2e$.
Since both are accelerated through the same potential difference $V$,the ratio of their wavelengths is:
$\frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2m_p q_p V}}}{\frac{h}{\sqrt{2m_{\alpha} q_{\alpha} V}}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}}$
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 $2\sqrt{2}: 1$.

(A) For a proton with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.
Thus,$p = \sqrt{2mE}$.
The de-Broglie wavelength of the proton is $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon with energy $E$,the wavelength $\lambda_2$ is given by $E = \frac{hc}{\lambda_2}$,which implies $\lambda_2 = \frac{hc}{E}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{1}{c} \sqrt{\frac{E}{2m}}$.
Therefore,$\frac{\lambda_1}{\lambda_2} \propto E^{1/2}$.
Comparing this with $\frac{\lambda_1}{\lambda_2} \propto E^n$,we get $n = \frac{1}{2}$.

(C) For an electron with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.
The de-Broglie wavelength of the electron is $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon,the energy $E$ is related to its wavelength $\lambda_p$ by $E = \frac{hc}{\lambda_p}$,which implies $\lambda_p = \frac{hc}{E}$.
The ratio of the wavelengths is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \sqrt{\frac{E}{2m}}$.

(A) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}} = \frac{1.228}{\sqrt{V}} \text{ nm}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial wavelength be $\lambda_1 = \lambda$ at potential $V_1 = V$.
Let the new wavelength be $\lambda_2$ at potential $V_2 = 4V$.
Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V}{4V}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,$\lambda_2 = \frac{\lambda}{2}$.
Thus,the wavelength reduces to half of its original value.

(A) The kinetic energy $K$ is related to momentum $p$ by the equation $K = \frac{p^2}{2m}$.
Therefore,$p = \sqrt{2mK}$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.
This implies $\lambda \propto \frac{1}{\sqrt{K}}$.
Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = 16K_1$.
Then,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{K_1}{16K_1}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
So,$\lambda_2 = 0.25 \lambda_1$.
The percentage change in wavelength is $\frac{\lambda_1 - \lambda_2}{\lambda_1} \times 100\% = \frac{\lambda_1 - 0.25 \lambda_1}{\lambda_1} \times 100\% = 0.75 \times 100\% = 75\%$.

(B) Given that the work function is negligible,the kinetic energy of the emitted electron is equal to the energy of the incident photon.
$\frac{1}{2}mv^2 = \frac{hc}{\lambda}$
Multiplying both sides by $2m$,we get $m^2v^2 = \frac{2mhc}{\lambda}$.
Taking the square root,the momentum $p = mv = \sqrt{\frac{2mhc}{\lambda}}$.
The de-Broglie wavelength $\lambda_e$ is given by $\lambda_e = \frac{h}{p}$.
Substituting the value of $p$:
$\lambda_e = \frac{h}{\sqrt{\frac{2mhc}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2mhc}} = \sqrt{\frac{h\lambda}{2mc}}$.

(D) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial potential difference be $V_1 = V$ and the initial wavelength be $\lambda_1$.
When the potential difference is doubled,the new potential difference is $V_2 = 2V$.
The new wavelength $\lambda_2$ is given by $\lambda_2 = \frac{h}{\sqrt{2me(2V)}} = \frac{1}{\sqrt{2}} \times \frac{h}{\sqrt{2meV}}$.
Therefore,$\lambda_2 = \frac{1}{\sqrt{2}} \lambda_1$.
Thus,the de-Broglie wavelength changes by a factor of $1/\sqrt{2}$.

(B) Both the electron and the proton are accelerated through the same potential difference $V$,so they acquire the same kinetic energy $K = eV$.
The momentum $P$ of a particle is related to its kinetic energy $K$ by the formula $P = \sqrt{2mK}$.
For the electron,the momentum is $P_{e} = \sqrt{2m_{e}K}$.
For the proton,the momentum is $P_{p} = \sqrt{2m_{p}K}$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{P}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_{p}}{\lambda_{e}} = \frac{h/P_{p}}{h/P_{e}} = \frac{P_{e}}{P_{p}}$.
Substituting the expressions for momentum: $\frac{\lambda_{p}}{\lambda_{e}} = \frac{\sqrt{2m_{e}K}}{\sqrt{2m_{p}K}} = \sqrt{\frac{m_{e}}{m_{p}}} = \left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}$.

(A) The de-Broglie wavelength $\lambda$ of an electron with energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.
From this relation,we can 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 $\frac{\lambda_1}{\lambda_2} = \frac{10^{-10}}{0.5 \times 10^{-10}} = 2$.
Substituting this into the ratio equation: $2 = \sqrt{\frac{E_2}{E_1}}$.
Squaring both sides,we get $4 = \frac{E_2}{E_1}$,so $E_2 = 4E_1 = 4E$.
The energy imparted to the electron is $\Delta E = E_2 - E_1 = 4E - E = 3E$.

(B) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ by the equation $\lambda = \frac{h}{p}$.
Since kinetic energy $K.E. = \frac{p^2}{2m}$,we can express momentum as $p = \sqrt{2m(K.E.)}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.
Squaring both sides,we obtain $\lambda^2 = \frac{h^2}{2m(K.E.)}$.
Rearranging the terms to solve for kinetic energy,we get $K.E. = \frac{h^2}{2m\lambda^2}$.

(C) The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.
From this,the kinetic energy $KE$ can be expressed as $KE = \frac{h^2}{2m\lambda^2}$.
Since the de-Broglie wavelength $\lambda$ is the same for both the electron and the proton,we have $KE \propto \frac{1}{m}$.
Because the mass of an electron $(m_e \approx 9.11 \times 10^{-31} \ kg)$ is much smaller than the mass of a proton $(m_p \approx 1.67 \times 10^{-27} \ kg)$,the kinetic energy of the electron will be greater than that of the proton.

(B) For a photon,the energy is given by $E = \frac{hc}{\lambda_p}$.
Therefore,$\lambda_p = \frac{hc}{E} \dots (i)$.
For a non-relativistic electron,the de-Broglie wavelength is $\lambda_e = \frac{h}{p}$.
Since $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Thus,$\lambda_e = \frac{h}{\sqrt{2mE}}$,which implies $E = \frac{h^2}{2m\lambda_e^2}$.
Substituting $E$ into equation $(i)$:
$\lambda_p = \frac{hc}{(h^2 / 2m\lambda_e^2)} = \frac{2mc}{h} \lambda_e^2$.
Therefore,$\lambda_p \propto \lambda_e^2$.

(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 an electron: $\lambda = \frac{h}{\sqrt{2meV}}$.
For a proton of mass $M$ and charge $e$ accelerated through potential $9V$: $\lambda' = \frac{h}{\sqrt{2Me(9V)}} = \frac{h}{3\sqrt{2MeV}}$.
Dividing $\lambda'$ by $\lambda$: $\frac{\lambda'}{\lambda} = \frac{\frac{h}{3\sqrt{2MeV}}}{\frac{h}{\sqrt{2meV}}} = \frac{1}{3} \sqrt{\frac{m}{M}}$.
Therefore,$\lambda' = \frac{\lambda}{3} \sqrt{\frac{m}{M}}$.

(B) The de Broglie wavelength $\lambda$ associated with a particle of mass $m$ and kinetic energy $E$ is given by the relation: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
From this expression,it is clear that $\lambda \propto \frac{1}{\sqrt{E}}$.
Therefore,when the kinetic energy $E$ of an electron is increased,the wavelength $\lambda$ will decrease.

(D) The momentum $p$ of a charged particle accelerated from rest through a potential difference $V$ is given by the relation $p = \sqrt{2mqV}$,where $m$ is the mass and $q$ is the charge of the particle.
For an electron,$m = m_{e}$ and $q = e$.
For an $\alpha$-particle,$m = m_{\alpha}$ and $q = 2e$.
Taking the ratio of the momenta:
$\frac{p_{e}}{p_{\alpha}} = \frac{\sqrt{2 m_{e} e V}}{\sqrt{2 m_{\alpha} (2e) V}}$
$\frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{2 m_{e} e V}{4 m_{\alpha} e V}}$
$\frac{p_{e}}{p_{\alpha}} = \sqrt{\frac{m_{e}}{2 m_{\alpha}}}$

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $m = 0.12 \ kg$
Velocity $v = 20 \ m \ s^{-1}$
Planck's constant $h = 6.63 \times 10^{-34} \ J \ s$
Substituting the values into the formula:
$\lambda = \frac{6.63 \times 10^{-34}}{0.12 \times 20}$
$\lambda = \frac{6.63 \times 10^{-34}}{2.4}$
$\lambda = 2.7625 \times 10^{-34} \ m$
Rounding to two decimal places,we get $\lambda = 2.76 \times 10^{-34} \ m$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.
Given values are $h = 6.63 \times 10^{-34} \ J \cdot s$,$m_{e} = 9.11 \times 10^{-31} \ kg$,and $v = 6.4 \times 10^{6} \ m/s$.
Substituting these values into the formula:
$\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 6.4 \times 10^{6}}$
$\lambda = \frac{6.63 \times 10^{-34}}{58.304 \times 10^{-25}}$
$\lambda \approx 0.1137 \times 10^{-9} \ m$
Since $1 \ nm = 10^{-9} \ m$,we get $\lambda \approx 0.114 \ nm$.

(D) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.
Rearranging the formula to solve for the speed $v$:
$v = \frac{h}{m \lambda}$.
Given values:
$h = 6.625 \times 10^{-34} \,J \,s$
$m = 1.0 \times 10^{-9} \,kg$
$\lambda = 3 \times 10^{-25} \,m$
Substituting these values into the equation:
$v = \frac{6.625 \times 10^{-34}}{(1.0 \times 10^{-9}) \times (3 \times 10^{-25})}$
$v = \frac{6.625 \times 10^{-34}}{3.0 \times 10^{-34}}$
$v = 2.2083... \,m \,s^{-1} \approx 2.2 \,m \,s^{-1}$.
Thus, the correct option is $D$.

(A) 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 matter waves.
The wavelength of these matter waves is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Because the wavelength of electrons is much smaller than that of visible light,electron microscopes can achieve much higher resolution than optical microscopes.
Therefore,the correct option is $(A)$.

(D) According to the de Broglie equation,the wavelength $\lambda$ of a particle is given by:
$\lambda = \frac{h}{mv}$
When a particle is dropped from a height $H$,its velocity $v$ at the ground is given by the equation of motion:
$v = \sqrt{2gH}$
Substituting the expression for velocity into the de Broglie equation:
$\lambda = \frac{h}{m\sqrt{2gH}}$
Since $h$,$m$,and $g$ are constants,we can write:
$\lambda \propto \frac{1}{\sqrt{H}}$
Therefore,the de Broglie wavelength depends on height as:
$\lambda \propto H^{-\frac{1}{2}}$