Matter Waves and de Broglie Wavelength Questions in English

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e K}}$.
Given $K = 9 \ eV$ and $m_e c^2 = 0.5 \ MeV = 0.5 \times 10^6 \ eV$.
We know $m_e = \frac{0.5 \times 10^6 \ eV}{c^2}$.
Substituting this into the wavelength formula:
$\lambda = \frac{h}{\sqrt{2 \left(\frac{0.5 \times 10^6}{c^2}\right) K}} = \frac{hc}{\sqrt{2 \times 0.5 \times 10^6 \times K}}$.
Given $h = 4 \times 10^{-15} \ eV \cdot s$ and $c = 3 \times 10^{10} \ cm/s$,then $hc = (4 \times 10^{-15}) \times (3 \times 10^{10}) = 12 \times 10^{-5} \ eV \cdot cm$.
Now,$\lambda = \frac{12 \times 10^{-5}}{\sqrt{2 \times 0.5 \times 10^6 \times 9}} = \frac{12 \times 10^{-5}}{\sqrt{9 \times 10^6}} = \frac{12 \times 10^{-5}}{3 \times 10^3} = 4 \times 10^{-8} \ cm$.

(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_p qV}}$.
Substituting the values: $h = 6.63 \times 10^{-34} \,J-s$, $m_p = 1.67 \times 10^{-27} \,kg$, $q = 1.6 \times 10^{-19} \,C$, and $V = 1000 \,V$.
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times 1000}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{5.344 \times 10^{-43}}} = \frac{6.63 \times 10^{-34}}{7.31 \times 10^{-22}} \approx 9.07 \times 10^{-13} \,m$.
Thus, the wavelength is approximately $9.1 \times 10^{-13} \,m$.

(D) We know that the de-Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
This implies that $K \propto \frac{1}{\lambda^2}$.
Let $K_1$ be the initial kinetic energy at $\lambda_1 = 1 \,nm$ and $K_2$ be the final kinetic energy at $\lambda_2 = 0.5 \,nm$.
Then, $\frac{K_2}{K_1} = \left( \frac{\lambda_1}{\lambda_2} \right)^2 = \left( \frac{1}{0.5} \right)^2 = 2^2 = 4$.
So, $K_2 = 4K_1$.
The energy that should be added to the electron is $\Delta K = K_2 - K_1 = 4K_1 - K_1 = 3K_1$.
Therefore, the energy to be added is three-times the initial energy.

(B) The de-Broglie wavelength $\lambda$ of a charged particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2mqV}}$
Since $h$,$m$,and $q$ are constants for the same particle,we have:
$\lambda \propto \frac{1}{\sqrt{V}}$
Therefore,the ratio of the wavelengths is:
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \frac{V_2^{1/2}}{V_1^{1/2}}$
This can be written as $V_2^{1/2} : V_1^{1/2}$.

(D) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Rearranging for energy $E$,we get $E = \frac{h^2}{2m\lambda^2}$.
Given values: $h = 6.6 \times 10^{-34} \ \text{Js}$,$m = 9.1 \times 10^{-31} \ \text{kg}$,and $\lambda = 5500 \ \text{Å} = 5.5 \times 10^{-7} \ \text{m}$.
Substituting these values into the formula:
$E = \frac{(6.6 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (5.5 \times 10^{-7})^2}$
$E = \frac{43.56 \times 10^{-68}}{18.2 \times 10^{-31} \times 30.25 \times 10^{-14}}$
$E = \frac{43.56 \times 10^{-68}}{550.55 \times 10^{-45}}$
$E \approx 0.0791 \times 10^{-23} \ \text{J} \approx 7.91 \times 10^{-25} \ \text{J}$.
Rounding to the nearest provided option,we get $8 \times 10^{-25} \ \text{J}$.

(B) Given that the energy of the photon $E$ is equal to the kinetic energy of the proton $K_p = E$.
For a proton,the de-Broglie wavelength $\lambda_1$ is given by $\lambda_1 = \frac{h}{p}$,where $p$ is the momentum of the proton.
Since $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Thus,$\lambda_1 = \frac{h}{\sqrt{2mE}}$.
For a photon,the wavelength $\lambda_2$ is given by $\lambda_2 = \frac{hc}{E}$.
Now,calculating the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} \cdot \frac{E}{hc} = \frac{E}{c \sqrt{2mE}} = \frac{\sqrt{E}}{c \sqrt{2m}}$.
Since $c$ and $m$ are constants,$\frac{\lambda_1}{\lambda_2} \propto \sqrt{E}$ or $E^{1/2}$.

(A) The de-Broglie wavelength $\lambda$ for a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For the proton $(p)$: $\lambda_p = \frac{h}{\sqrt{2m_p q_p V}}$.
For the $\alpha$-particle $(\alpha)$: $\lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}}$.
Given $\lambda_p = \lambda_\alpha$, we have $\sqrt{2m_p q_p V} = \sqrt{2m_\alpha q_\alpha V_\alpha}$.
Squaring both sides: $m_p q_p V = m_\alpha q_\alpha V_\alpha$.
We know that $m_\alpha = 4m_p$ and $q_\alpha = 2q_p$.
Substituting these values: $m_p q_p V = (4m_p)(2q_p) V_\alpha$.
$m_p q_p V = 8 m_p q_p V_\alpha$.
Therefore, $V_\alpha = \frac{V}{8}$.

(D) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{mv}$,where $v$ is the velocity of the electron.
The wavelength of a photon is given by $\lambda_p = \frac{h}{p_p} = \frac{hc}{E_p}$,where $E_p$ is the energy of the photon.
Given that $\lambda_e = \lambda_p$,we have $\frac{h}{mv} = \frac{hc}{E_p}$.
This implies $E_p = mvc$.
The kinetic energy of the electron is $K_e = \frac{1}{2}mv^2$.
The ratio of the kinetic energy of the electron to that of the photon is $\frac{K_e}{E_p} = \frac{\frac{1}{2}mv^2}{mvc} = \frac{v}{2c}$.
Substituting the given values $v = 1.5 \times 10^8 \ m/s$ and $c = 3 \times 10^8 \ m/s$:
$\frac{K_e}{E_p} = \frac{1.5 \times 10^8}{2 \times 3 \times 10^8} = \frac{1.5}{6} = \frac{1}{4}$.

(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 a proton,$\lambda_0 = \frac{h}{\sqrt{2m_p q_p V}}$.
For an $\alpha$-particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2q_p$.
Substituting these values into the formula for the $\alpha$-particle:
$\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(2q_p)V}} = \frac{h}{\sqrt{8(2m_p q_p V)}} = \frac{1}{\sqrt{8}} \left( \frac{h}{\sqrt{2m_p q_p V}} \right)$.
Since $\sqrt{8} = 2\sqrt{2}$,we get $\lambda_{\alpha} = \frac{\lambda_0}{2\sqrt{2}}$.

(D) Kinetic energy,$KE = 80 eV = 80 \times 1.6 \times 10^{-19} J = 128 \times 10^{-19} J$.
De-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 128 \times 10^{-19}}} = \frac{6.6 \times 10^{-34}}{\sqrt{256 \times 9 \times 10^{-50}}}$.
$\lambda = \frac{6.6 \times 10^{-34}}{16 \times 3 \times 10^{-25}} = \frac{6.6}{48} \times 10^{-9} m$.
$\lambda = 0.1375 \times 10^{-9} m \approx 1.375 \times 10^{-10} m = 1.375 Å$.
Rounding to the nearest value,we get $\lambda \approx 1.4 Å$.

(C) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_{e} K}}$, where $K$ is the kinetic energy.
Given: $h = 6.0 \times 10^{-34} \text{ J s}$, $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, $K = 320 \text{ eV} = 320 \times 1.6 \times 10^{-19} \text{ J}$.
Substituting the values:
$\lambda = \frac{6.0 \times 10^{-34}}{\sqrt{2 \times 9.0 \times 10^{-31} \times 320 \times 1.6 \times 10^{-19}}}$
$\lambda = \frac{6.0 \times 10^{-34}}{\sqrt{18 \times 10^{-31} \times 512 \times 10^{-19}}}$
$\lambda = \frac{6.0 \times 10^{-34}}{\sqrt{9216 \times 10^{-50}}}$
$\lambda = \frac{6.0 \times 10^{-34}}{96 \times 10^{-25}}$
$\lambda = 0.0625 \times 10^{-9} \text{ m} = 62.5 \times 10^{-12} \text{ m} = 62.5 \text{ pm}$.

(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.
Since kinetic energy $K = \frac{p^2}{2m}$,the momentum $p = \sqrt{2mK}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.
For a constant kinetic energy $K$,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.
Comparing the masses: $m_{\text{electron}} < m_{\text{proton}} \approx m_{\text{neutron}} < m_{\alpha\text{-particle}}$.
Since the $\alpha$-particle has the largest mass among the given options,it will have the shortest de-Broglie wavelength.

(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.
Since kinetic energy $K = \frac{p^2}{2m}$, we have $p = \sqrt{2mK}$.
Thus, $\lambda = \frac{h}{\sqrt{2mK}}$.
Given $h = 4.14 \times 10^{-15} eVs$, $K = 100 eV$, and $m = \frac{0.5 \times 10^6}{c^2} eV/c^2$.
Substituting these values, where $c = 3 \times 10^8 m/s$:
$\lambda = \frac{4.14 \times 10^{-15}}{\sqrt{2 \times (0.5 \times 10^6 / c^2) \times 100}}$
$\lambda = \frac{4.14 \times 10^{-15} \times c}{\sqrt{10^8}}$
$\lambda = \frac{4.14 \times 10^{-15} \times 3 \times 10^8}{10^4} = 1.242 \times 10^{-10} m = 124.2 pm$.

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Initially,let the mass be $m$,speed be $v$,and kinetic energy be $E = \frac{1}{2}mv^2$. Thus,$\lambda = \frac{h}{\sqrt{2mE}}$.
After some passengers leave,let the new mass be $m'$,new speed be $v' = 2v$,and new kinetic energy be $E' = 2E$.
Using the kinetic energy formula: $E' = \frac{1}{2}m'(v')^2$.
Substituting the values: $2E = \frac{1}{2}m'(2v)^2 = \frac{1}{2}m'(4v^2) = 2m'v^2$.
Since $E = \frac{1}{2}mv^2$,we have $2(\frac{1}{2}mv^2) = 2m'v^2$,which implies $m' = \frac{m}{2}$.
Now,the new de-Broglie wavelength $\lambda'$ is:
$\lambda' = \frac{h}{\sqrt{2m'E'}} = \frac{h}{\sqrt{2(\frac{m}{2})(2E)}} = \frac{h}{\sqrt{2mE}} = \lambda$.
Therefore,the new de-Broglie wavelength remains $\lambda$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $K$ is the kinetic energy.
Since the de-Broglie wavelength $\lambda$ is the same for all three particles,we have $\sqrt{2mK} = \text{constant}$,which implies $mK = \text{constant}$ or $K \propto \frac{1}{m}$.
The masses of the particles are $m_e$ (electron),$m_p$ (proton),and $m_{\alpha}$ (alpha particle). We know that $m_e < m_p < m_{\alpha}$.
Since $K$ is inversely proportional to mass,we have $\varepsilon_1 > \varepsilon_3 > \varepsilon_2$ (where $\varepsilon_1$ corresponds to the electron,$\varepsilon_3$ to the proton,and $\varepsilon_2$ to the alpha particle).

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.
Since the particles are accelerated by the same potential difference $V$,the kinetic energy $KE = qV$.
Thus,$\lambda = \frac{h}{\sqrt{2mqV}}$.
Since $h$,$q$,and $V$ are constants for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{\lambda_{p}}{\lambda_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$.
Given $m_{p} = 2000 m_{e}$,we substitute this into the ratio:
$\frac{\lambda_{p}}{\lambda_{e}} = \sqrt{\frac{m_{e}}{2000 m_{e}}} = \sqrt{\frac{1}{2000}} = \frac{1}{\sqrt{400 \times 5}} = \frac{1}{20 \sqrt{5}}$.
Hence,$\lambda_{p} = \frac{\lambda_{e}}{20 \sqrt{5}}$.

(D) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2 m_e e V}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Therefore,$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.
Given that $\lambda_1 = \lambda$ and $\lambda_2 = 2\lambda$,we have $\frac{\lambda}{2\lambda} = \sqrt{\frac{V_2}{10000}}$.
Squaring both sides,we get $\frac{1}{4} = \frac{V_2}{10000}$.
Thus,$V_2 = \frac{10000}{4} = 2500 \ V$.

(D) The de-Broglie wavelength $\lambda$ is related to kinetic energy $E$ by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
From this,we can see that $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_2}{E_1}}$.
Given:
$\lambda_1 = 0.4 \times 10^{-10} \ m$
$E_1 = 1.0 \ keV$
$\lambda_2 = 1.0 \times 10^{-10} \ m$
Substituting these values into the ratio:
$\frac{0.4 \times 10^{-10}}{1.0 \times 10^{-10}} = \sqrt{\frac{E_2}{1.0 \ keV}}$
$0.4 = \sqrt{E_2}$
Squaring both sides:
$E_2 = (0.4)^2 = 0.16 \ keV$.

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2meV}}$
Squaring both sides, we get:
$\lambda^2 = \frac{h^2}{2meV}$
Rearranging for $V$:
$V = \frac{h^2}{2me\lambda^2}$
Using the standard approximation formula for an electron:
$V \approx \frac{150}{\lambda^2} \text{ V}$, where $\lambda$ is in $\text{Å}$.
Given $\lambda = 1 \text{ Å}$:
$V = \frac{150}{(1)^2} = 150 \text{ V}$.
Thus, the required potential difference is $150 \text{ V}$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Case $A$: For free fall,$v = \sqrt{2gh}$. Since $v$ is independent of mass,$\lambda = \frac{h}{m\sqrt{2gh}} \propto \frac{1}{m}$. Thus,the heavier particle has a smaller wavelength.
Case $B$: With same kinetic energy $K$,$\lambda = \frac{h}{\sqrt{2mK}}$. Since $\lambda \propto \frac{1}{\sqrt{m}}$,the heavier particle has a smaller wavelength.
Case $C$: With same momentum $p$,$\lambda = \frac{h}{p}$. Since $p$ is the same,$\lambda$ is the same for both.
Case $D$: With same speed $v$,$\lambda = \frac{h}{mv}$. Since $\lambda \propto \frac{1}{m}$,the heavier particle has a smaller wavelength.
Note: Options $A$,$B$,and $D$ all result in the heavier particle having a smaller de-Broglie wavelength. However,in standard physics problems of this type,the most direct relationship is often associated with constant speed or constant kinetic energy. Given the options,$D$ is the most fundamental case where $\lambda \propto 1/m$ is directly observed.

(B) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
Given values are:
$h = 6.6 \times 10^{-34} \ J \cdot s$
$m = 1 \times 10^{-30} \ kg$
$K = 200 \ eV = 200 \times 1.6 \times 10^{-19} \ J = 3.2 \times 10^{-17} \ J$.
Substituting these values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (1 \times 10^{-30}) \times (3.2 \times 10^{-17})}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{6.4 \times 10^{-47}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{8 \times 10^{-23.5}}$
Using the calculation $\sqrt{64 \times 10^{-48}} = 8 \times 10^{-24}$,we get:
$\lambda = \frac{6.6 \times 10^{-34}}{8 \times 10^{-24}} = 0.825 \times 10^{-10} \ m = 8.25 \times 10^{-11} \ m$.

(A) The de-Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{m_e v_e}$,where $v_e = c/2$. So,$\lambda_e = \frac{h}{m_e (c/2)} = \frac{2h}{m_e c}$.
For a photon,the wavelength is $\lambda_p = \frac{h}{p_p} = \frac{h}{E_p/c} = \frac{hc}{E_p}$,where $E_p$ is the energy of the photon.
Given $\lambda_e = \lambda_p$,we have $\frac{2h}{m_e c} = \frac{hc}{E_p}$.
This implies $E_p = \frac{m_e c^2}{2}$.
The kinetic energy of the electron is $K_e = \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_e (c/2)^2 = \frac{1}{8} m_e c^2$.
The ratio of kinetic energies is $\frac{K_e}{E_p} = \frac{\frac{1}{8} m_e c^2}{\frac{1}{2} m_e c^2} = \frac{1}{4}$.

(C) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK_e}}$,where $K_e$ is the kinetic energy of the electron and $m$ is its mass.
Thus,$K_e = \frac{h^2}{2m\lambda^2}$.
The energy of a photon is given by $E_p = \frac{hc}{\lambda}$.
We are given that the wavelengths are equal,so $\lambda_e = \lambda_p = \lambda$.
The ratio of the energy of the photon to the kinetic energy of the electron is:
$\frac{E_p}{K_e} = \frac{hc/\lambda}{h^2/(2m\lambda^2)} = \frac{hc}{\lambda} \cdot \frac{2m\lambda^2}{h^2} = \frac{2mc\lambda}{h}$.
Since $E_p = \frac{hc}{\lambda}$,we have $\lambda = \frac{hc}{E_p}$.
Substituting this into the ratio:
$\frac{E_p}{K_e} = \frac{2mc}{h} \cdot \frac{hc}{E_p} = \frac{2mc^2}{E_p}$.
Given $mc^2 = 0.5 \ MeV = 500 \ keV$ and $E_p = 50 \ keV$:
$\frac{E_p}{K_e} = \frac{2 \times 500 \ keV}{50 \ keV} = \frac{1000}{50} = 20$.
Therefore,the ratio is $20:1$.

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$.
For a deuteron $(d)$,the mass $m_d = 2m_p$ and kinetic energy $KE_d = E$.
For an alpha particle $(\alpha)$,the mass $m_{\alpha} = 4m_p$ and kinetic energy $KE_{\alpha} = 2E$.
The ratio of wavelengths is $\frac{\lambda_d}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} \cdot KE_{\alpha}}{m_d \cdot KE_d}}$.
Substituting the values: $\frac{\lambda_d}{\lambda_{\alpha}} = \sqrt{\frac{4m_p \cdot 2E}{2m_p \cdot E}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$.
Thus,the ratio is $2:1$,which means $n = 2$.

(C) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.
Given: $h = 6.6 \times 10^{-34} \text{ J} \cdot \text{s}$,$q = 3 \times 10^{-19} \text{ C}$,$m = 6 \times 10^{-27} \text{ kg}$,and $V = 1.21 \text{ V}$.
Substituting the values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (6 \times 10^{-27}) \times (3 \times 10^{-19}) \times 1.21}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{36 \times 10^{-46} \times 1.21}}$
$\lambda = \frac{6.6 \times 10^{-34}}{6 \times 10^{-23} \times 1.1}$
$\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}}$
$\lambda = 10^{-11} \text{ m} = 10 \times 10^{-12} \text{ m}$.
Comparing this with $\alpha \times 10^{-12} \text{ m}$,we get $\alpha = 10$.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.
For a gas molecule,the average kinetic energy is $K = \frac{3}{2}kT$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2m(\frac{3}{2}kT)}} = \frac{h}{\sqrt{3mkT}}$.
Given values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 5.31 \times 10^{-26} \ kg$,$k = 1.38 \times 10^{-23} \ J/K$,and $T = 27 + 273 = 300 \ K$.
Calculating the denominator: $\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300} = \sqrt{658.734 \times 10^{-49}} = \sqrt{65.8734 \times 10^{-48}} \approx 8.116 \times 10^{-24}$.
$\lambda = \frac{6.63 \times 10^{-34}}{8.116 \times 10^{-24}} \approx 0.8168 \times 10^{-10} \ m = 81.68 \times 10^{-12} \ m$.
Wait,re-calculating: $\sqrt{3 \times 5.31 \times 1.38 \times 300} \times 10^{-24.5} \approx \sqrt{6587.34} \times 10^{-24.5} \approx 81.16 \times 10^{-24.5}$.
Correct calculation: $\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}} = \frac{6.63 \times 10^{-34}}{\sqrt{6587.34 \times 10^{-49}}} = \frac{6.63 \times 10^{-34}}{2.566 \times 10^{-23}} \approx 2.58 \times 10^{-11} \ m = 25.8 \times 10^{-12} \ m$.
Rounding to the nearest integer,$x = 26$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given:
Mass $m = 0.033 \text{ kg}$
Velocity $v = 1 \text{ km/s} = 1000 \text{ m/s} = 10^3 \text{ m/s}$
Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$
Substituting the values into the formula:
$\lambda = \frac{6.6 \times 10^{-34}}{0.033 \times 1000}$
$\lambda = \frac{6.6 \times 10^{-34}}{33}$
$\lambda = 0.2 \times 10^{-34} \text{ m}$
$\lambda = 2 \times 10^{-35} \text{ m}$.

(B) For a photon,the momentum $p$ is given by $p = \frac{h}{\lambda}$,where $h$ is Planck's constant.
According to the de Broglie relation,the wavelength $\lambda'$ associated with a particle of momentum $p$ is $\lambda' = \frac{h}{p}$.
Substituting the expression for $p$ into the de Broglie equation,we get $\lambda' = \frac{h}{h/\lambda} = \lambda$.
Therefore,the de Broglie wavelength of a photon is equal to the wavelength of the electromagnetic radiation associated with it.
Thus,the correct option is $B$.

(A) The force on the electron is $\vec{F} = q\vec{E} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$.
The acceleration of the electron is $a = \frac{F}{m} = \frac{2eE_0}{m}$.
The velocity of the electron at time $t$ is given by $v(t) = v_0 + at = v_0 + \left(\frac{2eE_0}{m}\right)t = v_0 \left[1 + \frac{2eE_0 t}{m v_0}\right]$.
The de Broglie wavelength at time $t$ is $\lambda = \frac{h}{mv(t)}$.
Substituting $v(t)$,we get $\lambda = \frac{h}{m v_0 [1 + \frac{2eE_0 t}{m v_0}]} = \frac{\lambda_0}{[1 + \frac{2eE_0 t}{m v_0}]}$.

(C) In free space,the de Broglie wavelength is given by $\lambda_0 = h/mv$.
When the electron enters the medium,its velocity is reduced by $20\%$.
Therefore,the new velocity $v' = v - 0.20v = 0.8v$.
The new de Broglie wavelength in the medium is $\lambda = h/mv'$.
Substituting $v' = 0.8v$,we get $\lambda = h/(m \times 0.8v) = (1/0.8) \times (h/mv)$.
Since $\lambda_0 = h/mv$,we have $\lambda = (1/0.8) \times \lambda_0 = 1.25 \lambda_0$.
Comparing this with $\alpha\lambda_0$,we find that $\alpha = 1.25$.

(A) The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$.
Since both the electron and the proton are accelerated through the same potential difference $V$ and have the same magnitude of charge $q$,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,the ratio of the de Broglie wavelengths is $\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}$.

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.
This implies $\lambda \propto \frac{1}{\sqrt{V}}$.
Let $\lambda_1$ be the wavelength at potential $V_1$,so $\lambda_1 = \frac{k}{\sqrt{V_1}}$,where $k$ is a constant.
When the potential is changed to $V_2$,the wavelength increases by $50\%$,so $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1$.
Thus,$\lambda_2 = \frac{k}{\sqrt{V_2}}$.
Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{\sqrt{V_1}}{\sqrt{V_2}} = 1.5$.
Squaring both sides: $\frac{V_1}{V_2} = (1.5)^2 = 2.25$.
We can write $2.25$ as $\frac{225}{100} = \frac{9}{4}$.
Given $(V_1/V_2) = (9/\alpha)$,comparing the two expressions gives $\alpha = 4$.